Discrete and Continuous Probability Distributions

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Introduction to Probability Distributions

A probability distribution is a mathematical description that defines the likelihood of different outcomes in a random process. It tells us the probabilities of various possible outcomes of a random variable.

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Difference Between Discrete and Continuous Distributions

1. Discrete Probability Distribution:
   • Deals with discrete random variables that take on a finite or countable set of values.
   • Examples include the number of heads in a series of coin flips or the number of cars arriving at a toll booth in an hour.
2. Continuous Probability Distribution:
   • Deals with continuous random variables that take an infinite number of possible values within a given range.
   • Examples include the height of people, time taken to finish a task, or temperature readings.

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Examples of Discrete and Continuous Variables

• Discrete Variables:
  • Number of students in a classroom
  • Number of cars in a parking lot
• Continuous Variables:
  • Weight of a package
  • Time taken to run a marathon

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Discrete Probability Distributions

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Properties of Discrete Distributions

• The probability of each individual outcome is a non-negative number.
• The sum of all probabilities for all possible outcomes is 1.

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Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials, where each trial results in either a success or failure.

Formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

Where:

• $n$ = number of trials
• $k$ = number of successes
• $p$ = probability of success in each trial

Example:
What is the probability of getting exactly 3 heads in 5 flips of a fair coin?

Answer:
Step 1: Given Data:
$n=5,k=3,p=\frac{1}{2}$
Step 2: Solution:
$P(X=3)=(\genfrac{}{}{0ex}{}{5}{3}){\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2$
Step 3: Final Answer:
$P(X=3)=\frac{10}{32}=0.3125$

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Geometric Distribution

The geometric distribution models the number of trials until the first success in a series of independent Bernoulli trials.

Formula:
$P(X=k)=(1–p{)}^{k-1}p$

Where:

• $p$ = probability of success
• $k$ = trial on which the first success occurs

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Poisson Distribution

The Poisson distribution models the number of occurrences of an event in a fixed interval of time or space.

Formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

Where:

• $\lambda$ = average rate of occurrence
• $k$ = number of occurrences

Example:
A call center receives an average of 5 calls per minute. What is the probability that they receive exactly 3 calls in the next minute?

Answer:
Step 1: Given Data:
$\lambda =5,k=3$
Step 2: Solution:
$P(X=3)=\frac{5^3{e}^{-5}}{3!}=\frac{125e^{-5}}{6}$
Step 3: Final Answer:
$P(X=3)\approx 0.1404$

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Hypergeometric Distribution

The hypergeometric distribution models the probability of $k$ successes in $n$ draws from a finite population of size $N$ without replacement.

Formula:
$P(X=k)=\frac{(\genfrac{}{}{0ex}{}{K}{k})(\genfrac{}{}{0ex}{}{N-K}{n-k})}{(\genfrac{}{}{0ex}{}{N}{n})}$

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Multinomial Distribution

The multinomial distribution generalizes the binomial distribution by allowing more than two possible outcomes.

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Continuous Probability Distributions

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Properties of Continuous Distributions

• Probabilities are defined over intervals, not individual outcomes.
• The probability density function (PDF) gives the likelihood of a random variable falling within a certain range.
• The area under the PDF curve for a specific interval gives the probability of the random variable being in that interval.

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Normal Distribution

The normal distribution is a continuous probability distribution that is symmetric around its mean.

Formula:
$f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x–\mu {)}^2}{2{\sigma }^2}}$

Where:

• $\mu$ = mean
• $\sigma$ = standard deviation
• $x$ = random variable

Example:
What is the probability that a normally distributed variable with mean 50 and standard deviation 10 is less than 60?

Answer:
Step 1: Given Data:
$\mu =50,\sigma =10,X=60$
Step 2: Solution:
Standardize the variable:
$Z=\frac{60–50}{10}=1$
Step 3: Final Answer:
$P(X<60)=P(Z<1)=0.8413$

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Uniform Distribution

The uniform distribution describes a situation where all outcomes are equally likely within a given interval.

Formula:
$f(x)=\frac{1}{b–a}\text{ for }a\le x\le b$

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Exponential Distribution

The exponential distribution models the time between events in a Poisson process.

Formula:
$f(x)=\lambda e^{-\lambda x}\text{ for }x\ge 0$

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Gamma Distribution

The gamma distribution generalizes the exponential distribution for the time to multiple events.

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Beta Distribution

The beta distribution is used to model random variables that are bounded between 0 and 1.

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Weibull Distribution

The Weibull distribution is used in reliability analysis and survival studies.

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Probability Density Function (PDF)

The PDF gives the likelihood of a continuous random variable taking on a specific value.

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Cumulative Distribution Function (CDF)

The CDF gives the probability that a random variable is less than or equal to a particular value.

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Moment Generating Functions for Distributions

Moment generating functions (MGF) are used to find the moments (mean, variance) of a distribution.

Formula:
$M_X(t)=E(e^{tX})$

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Expected Value and Variance for Discrete and Continuous Distributions

1. Expected Value (Discrete):
   $E(X)=\sum x_{i}P(x_i)$
2. Variance (Discrete):
   $\text{Var}(X)=E(X^2)–(E(X){)}^2$
3. Expected Value (Continuous):
   $E(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x),dx$
4. Variance (Continuous):
   $\text{Var}(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x–\mu {)}^{2}f(x),dx$

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Applications of Discrete and Continuous Distributions

• Discrete Distributions:
  • Modeling the number of customer arrivals at a store.
  • Modeling the number of defective products in a batch.
• Continuous Distributions:
  • Modeling the amount of rainfall in a year.
  • Modeling the time until a machine fails.

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Fitting Data to Distributions

To fit data to a distribution, we estimate the parameters of the distribution (e.g., mean and standard deviation for a normal distribution) and compare the theoretical distribution to the observed data.

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Using Distributions in Real-World Problems

• Finance: Modeling stock returns using normal distributions.
• Insurance: Using Poisson distributions to model claim counts.
• Engineering: Using Weibull distributions for reliability analysis.

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FAQs

1. What is the difference between discrete and continuous probability distributions?
   • Discrete distributions deal with countable outcomes, while continuous distributions deal with uncountable outcomes over an interval.
2. When should I use the binomial distribution?
   • Use the binomial distribution when you have a fixed number of independent trials with two possible outcomes (success or failure).
3. What is the significance of the normal distribution?
   • The normal distribution is significant because it describes many natural phenomena and is used in the central limit theorem.
4. How do you calculate the expected value for a continuous distribution?
   • The expected value for a continuous distribution is calculated using the integral:
     $E(X)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x),dx$
5. Why are distributions important in probability theory?
   • Distributions provide a structured way to model and analyze the likelihood of different outcomes and are essential for understanding uncertainty.

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Question And Answer Library

Example 1: Probability of a Single Event

Problem:
A fair six-sided die is rolled. What is the probability of rolling a 4?

Answer:

Step 1: Given Data:
Total outcomes = 6,
Favorable outcomes = 1 (rolling a 4).

Step 2: Solution:
Using the probability formula:
$P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(4)=\frac{1}{6}$.

Step 3: Final Answer:
$P(4)=\frac{1}{6}$.

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Example 2: Binomial Distribution

Problem:
In 10 flips of a coin, what is the probability of getting exactly 5 heads?

Answer:

Step 1: Given Data:
$n=10$,
$k=5$,
$p=0.5$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=5)=(\genfrac{}{}{0ex}{}{10}{5})(0.5{)}^5(0.5{)}^5$.

Calculate:
$(\genfrac{}{}{0ex}{}{10}{5})=252$.

So:
$P(X=5)=252\cdot {\left(\frac{1}{2}\right)}^{10}=252\cdot \frac{1}{1024}\approx 0.2461$.

Step 3: Final Answer:
$P(X=5)\approx 0.2461$.

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Example 3: Poisson Distribution

Problem:
A bookstore receives an average of 3 customers per hour. What is the probability that exactly 4 customers arrive in the next hour?

Answer:

Step 1: Given Data:
$\lambda =3$,
$k=4$.

Step 2: Solution:
Using the Poisson formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

$P(X=4)=\frac{3^4{e}^{-3}}{4!}$.

Calculate:
$=\frac{81e^{-3}}{24}\approx \frac{81\cdot 0.0498}{24}\approx 0.1680$.

Step 3: Final Answer:
$P(X=4)\approx 0.1680$.

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Example 4: Normal Distribution

Problem:
Find the probability that a randomly selected student scores less than 70 on a test that has a mean of 75 and a standard deviation of 10.

Answer:

Step 1: Given Data:
Mean $\mu =75$,
Standard deviation $\sigma =10$,
Score $X=70$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{X–\mu }{\sigma }=\frac{70–75}{10}=-0.5$.

Use z-tables to find:
$P(Z<-0.5)\approx 0.3085$.

Step 3: Final Answer:
$P(X<70)\approx 0.3085$.

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Example 5: Uniform Distribution

Problem:
If a random variable is uniformly distributed between 2 and 10, what is the probability that it falls between 3 and 7?

Answer:

Step 1: Given Data:
Lower bound $a=2$,
Upper bound $b=10$.

Step 2: Solution:
Calculate total range:
$Range=b–a=10–2=8$.

Calculate specific range:
$Rang{e}_{3-7}=7–3=4$.

Now calculate probability:
$P(3<X<7)=\frac{Rang{e}_{3-7}}{Range}=\frac{4}{8}=0.5$.

Step 3: Final Answer:
$P(3<X<7)=0.5$.

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Example 6: Expected Value for Discrete Variables

Problem:
For the random variable $X$ with values 1, 2, and 3, and probabilities $P(1)=0.2$, $P(2)=0.5$, $P(3)=0.3$, find the expected value.

Answer:

Step 1: Given Data:
Values: $1,2,3$
Probabilities: $0.2,0.5,0.3$.

Step 2: Solution:
Calculate expected value:
$E(X)=(1\cdot 0.2)+(2\cdot 0.5)+(3\cdot 0.3)$.

$E(X)=0.2+1+0.9=2.1$.

Step 3: Final Answer:
$E(X)=2.1$.

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Example 7: Variance of a Discrete Random Variable

Problem:
Given $P(1)=0.1$, $P(2)=0.2$, and $P(3)=0.7$, find the variance.

Answer:

Step 1: Given Data:
Probabilities: $0.1,0.2,0.7$.

Step 2: Solution:
Calculate expected value:
$E(X)=(1\cdot 0.1)+(2\cdot 0.2)+(3\cdot 0.7)$.

$E(X)=0.1+0.4+2.1=2.6$.

Calculate $E(X^2)$:
$E(X^2)=(1^2\cdot 0.1)+(2^2\cdot 0.2)+(3^2\cdot 0.7)$.

$E(X^2)=0.1+0.8+6.3=7.2$.

Calculate variance:
$Var(X)=E(X^2)–(E(X){)}^2=7.2–(2.6{)}^2=7.2–6.76=0.44$.

Step 3: Final Answer:
$Var(X)=0.44$.

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Example 8: Cumulative Distribution Function

Problem:
For the random variable $X$ with values $1,2,3$ and probabilities $P(1)=0.2$, $P(2)=0.5$, $P(3)=0.3$, find the CDF.

Answer:

Step 1: Given Data:
Probabilities: $P(1)=0.2$, $P(2)=0.5$, $P(3)=0.3$.

Step 2: Solution:
Calculate CDF:
$F(1)=P(X\le 1)=0.2$.

$F(2)=P(X\le 2)=0.2+0.5=0.7$.

$F(3)=P(X\le 3)=0.2+0.5+0.3=1$.

Step 3: Final Answer:
CDF values are:
$F(1)=0.2$,
$F(2)=0.7$,
$F(3)=1$.

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Example 9: Joint Probability

Problem:
In a survey, 40% prefer tea, 30% prefer coffee, and 10% prefer both. What is the probability of a person preferring tea or coffee?

Answer:

Step 1: Given Data:
$P(\text{Tea})=0.4$,
$P(\text{Coffee})=0.3$,
$P(\text{Tea and Coffee})=0.1$.

Step 2: Solution:
Calculate probability:
$P(\text{Tea or Coffee})=P(\text{Tea})+P(\text{Coffee})–P(\text{Tea and Coffee})$

$=0.4+0.3–0.1=0.6$.

Step 3: Final Answer:
$P(\text{Tea or Coffee})=0.6$.

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Example 10: Probability of Normal Distribution

Problem:
Find the probability that a randomly selected value from a normal distribution with mean 100 and standard deviation 15 is greater than 120.

Answer:

Step 1: Given Data:
Mean $\mu =100$,
Standard deviation $\sigma =15$,
Value $X=120$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{X–\mu }{\sigma }=\frac{120–100}{15}=\frac{20}{15}\approx 1.33$.

Use z-tables to find:
$P(Z>1.33)\approx 1–P(Z<1.33)\approx 1–0.9082=0.0918$.

Step 3: Final Answer:
$P(X>120)\approx 0.0918$.

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Example 11: Geometric Distribution

Problem:
What is the probability that the first success occurs on the 4th trial when the probability of success is 0.3?

Answer:

Step 1: Given Data:
$p=0.3$,
$k=4$.

Step 2: Solution:
Using the geometric distribution formula:
$P(X=k)=(1–p{)}^{k-1}p$

$P(X=4)=(0.7{)}^3(0.3)$.

Calculate:
$P(X=4)=0.343\cdot 0.3=0.1029$.

Step 3: Final Answer:
$P(X=4)\approx 0.1029$.

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Example 12: Continuous Uniform Distribution

Problem:
If a random variable is uniformly distributed between 10 and 20, what is the probability that it falls between 12 and 15?

Answer:

Step 1: Given Data:
Lower bound $a=10$,
Upper bound $b=20$.

Step 2: Solution:
Calculate total range:
$Range=b–a=20–10=10$.

Calculate specific range:
$Rang{e}_{12-15}=15–12=3$.

Now calculate probability:
$P(12<X<15)=\frac{Rang{e}_{12-15}}{Range}=\frac{3}{10}=0.3$.

Step 3: Final Answer:
$P(12<X<15)=0.3$.

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Example 13: Finding Expected Value for Continuous Distribution

Problem:
Find the expected value of a continuous uniform distribution from 5 to 15.

Answer:

Step 1: Given Data:
Lower bound $a=5$,
Upper bound $b=15$.

Step 2: Solution:
Using the formula for expected value of a uniform distribution:
$E(X)=\frac{a+b}{2}$

$E(X)=\frac{5+15}{2}=\frac{20}{2}=10$.

Step 3: Final Answer:
$E(X)=10$.

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Example 14: Finding Variance of Normal Distribution

Problem:
If a normal distribution has a mean of 50 and a standard deviation of 5, what is the variance?

Answer:

Step 1: Given Data:
Mean $\mu =50$,
Standard deviation $\sigma =5$.

Step 2: Solution:
Calculate variance:
$Var(X)={\sigma }^2=5^2=25$.

Step 3: Final Answer:
$Var(X)=25$.

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Example 15: Probability of Binomial Distribution

Problem:
A coin is flipped 6 times. What is the probability of getting exactly 2 tails?

Answer:

Step 1: Given Data:
$n=6$,
$k=2$,
$p=0.5$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=2)=(\genfrac{}{}{0ex}{}{6}{2})(0.5{)}^2(0.5{)}^4$.

Calculate:
$(\genfrac{}{}{0ex}{}{6}{2})=15$.

So:
$P(X=2)=15\cdot {\left(\frac{1}{2}\right)}^6=15\cdot \frac{1}{64}\approx 0.2344$.

Step 3: Final Answer:
$P(X=2)\approx 0.2344$.

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Example 16: Conditional Probability

Problem:
In a deck of cards, what is the probability of drawing a King given that the card drawn is a face card?

Answer:

Step 1: Given Data:
Total face cards = 12 (Jack, Queen, King for each suit).
Favorable outcomes = 4 (Kings).

Step 2: Solution:
Using conditional probability:
$P(King|Face)=\frac{P(King\cap Face)}{P(Face)}$

$P(King|Face)=\frac{4/52}{12/52}=\frac{4}{12}=\frac{1}{3}$.

Step 3: Final Answer:
$P(King|Face)=\frac{1}{3}$.

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Example 17: Probability of Exceeding a Value in Normal Distribution

Problem:
For a normal distribution with $\mu =100$ and $\sigma =15$, what is the probability that a randomly selected value is greater than 120?

Answer:

Step 1: Given Data:
Mean $\mu =100$,
Standard deviation $\sigma =15$,
Value $X=120$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{X–\mu }{\sigma }=\frac{120–100}{15}=\frac{20}{15}\approx 1.33$.

Use z-tables to find:
$P(Z>1.33)\approx 1–P(Z<1.33)\approx 1–0.9082=0.0918$.

Step 3: Final Answer:
$P(X>120)\approx 0.0918$.

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Example 18: Expected Value for Binomial Distribution

Problem:
Find the expected value of a binomial distribution with $n=8$ and $p=0.5$.

Answer:

Step 1: Given Data:
$n=8$,
$p=0.5$.

Step 2: Solution:
Using the expected value formula for binomial distribution:
$E(X)=n\cdot p$

$E(X)=8\cdot 0.5=4$.

Step 3: Final Answer:
$E(X)=4$.

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Example 19: Variance of a Binomial Distribution

Problem:
What is the variance of a binomial distribution with $n=10$ and $p=0.3$?

Answer:

Step 1: Given Data:
$n=10$,
$p=0.3$.

Step 2: Solution:
Using the variance formula for binomial distribution:
$Var(X)=n\cdot p\cdot (1–p)$

$Var(X)=10\cdot 0.3\cdot (1–0.3)$.

Calculate:
$Var(X)=10\cdot 0.3\cdot 0.7=2.1$.

Step 3: Final Answer:
$Var(X)=2.1$.

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Example 20: Probability from Cumulative Distribution Function

Problem:
For a continuous random variable $X$ with CDF $F(x)=0.5$ for $x=10$, what is the probability that $X$ is less than 10?

Answer:

Step 1: Given Data:
$F(10)=0.5$.

Step 2: Solution:
$P(X<10)=F(10)$.

Step 3: Final Answer:
$P(X<10)=0.5$.

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Example 21: Finding the Median of a Continuous Distribution

Problem:
For a continuous uniform distribution between 1 and 9, what is the median?

Answer:

Step 1: Given Data:
Lower bound $a=1$,
Upper bound $b=9$.

Step 2: Solution:
Using the median formula for uniform distribution:
$Median=\frac{a+b}{2}$

$Median=\frac{1+9}{2}=5$.

Step 3: Final Answer:
$Median=5$.

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Example 22: Joint Probability of Independent Events

Problem:
What is the probability of rolling a 3 on a die and flipping a heads on a coin?

Answer:

Step 1: Given Data:
$P(\text{Rolling a 3})=\frac{1}{6}$,
$P(\text{Flipping a Heads})=\frac{1}{2}$.

Step 2: Solution:
Since the events are independent:
$P(\text{Rolling a 3 and Heads})=P(\text{Rolling a 3})\cdot P(\text{Flipping a Heads})$

$=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$.

Step 3: Final Answer:
$P(\text{Rolling a 3 and Heads})=\frac{1}{12}$.

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Example 23: Variance of Poisson Distribution

Problem:
If the average rate of occurrence $\lambda$ for a Poisson distribution is 4, what is the variance?

Answer:

Step 1: Given Data:
$\lambda =4$.

Step 2: Solution:
Using the formula for variance of Poisson distribution:
$Var(X)=\lambda$

$Var(X)=4$.

Step 3: Final Answer:
$Var(X)=4$.

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Example 24: Probability of Multiple Events

Problem:
What is the probability of rolling either a 1 or a 2 on a fair six-sided die?

Answer:

Step 1: Given Data:
$P(\text{Rolling a 1})=\frac{1}{6}$,
$P(\text{Rolling a 2})=\frac{1}{6}$.

Step 2: Solution:
Since the events are mutually exclusive:
$P(\text{Rolling a 1 or 2})=P(\text{Rolling a 1})+P(\text{Rolling a 2})$

$=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$.

Step 3: Final Answer:
$P(\text{Rolling a 1 or 2})=\frac{1}{3}$.

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Example 25: CDF of Normal Distribution

Problem:
Find the probability that a normally distributed random variable with $\mu =30$ and $\sigma =5$ is between 25 and 35.

Answer:

Step 1: Given Data:
Mean $\mu =30$,
Standard deviation $\sigma =5$,
Values: $X_1=25$, $X_2=35$.

Step 2: Solution:
Calculate z-scores:
$Z_1=\frac{25–30}{5}=-1$,
$Z_2=\frac{35–30}{5}=1$.

Find probabilities:
$P(Z<1)\approx 0.8413$,
$P(Z<-1)\approx 0.1587$.

Calculate probability:
$P(25<X<35)=P(Z<1)–P(Z<-1)$

$=0.8413–0.1587=0.6826$.

Step 3: Final Answer:
$P(25<X<35)\approx 0.6826$.

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Example 26: Finding Probability Using Central Limit Theorem

Problem:
A sample of size 30 is taken from a population with a mean of 50 and a standard deviation of 10. What is the probability that the sample mean is greater than 53?

Answer:

Step 1: Given Data:
Population mean $\mu =50$,
Population standard deviation $\sigma =10$,
Sample size $n=30$.

Step 2: Solution:
Calculate the standard error:
$SE=\frac{\sigma }{\sqrt{n}}=\frac{10}{\sqrt{30}}\approx 1.8257$.

Calculate z-score:
$Z=\frac{53–50}{SE}=\frac{3}{1.8257}\approx 1.64$.

Use z-tables to find:
$P(Z>1.64)\approx 1–0.9495=0.0505$.

Step 3: Final Answer:
$P(\text{Sample mean}>53)\approx 0.0505$.

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Example 27: Mode of a Probability Distribution

Problem:
For the discrete probability distribution given by $P(1)=0.1$, $P(2)=0.4$, $P(3)=0.4$, and $P(4)=0.1$, find the mode.

Answer:

Step 1: Given Data:
Probabilities: $P(1)=0.1$, $P(2)=0.4$, $P(3)=0.4$, $P(4)=0.1$.

Step 2: Solution:
Identify the value with the highest probability:
The mode is 2 and 3 (both have probabilities of 0.4).

Step 3: Final Answer:
The mode is $2$ and $3$ (bimodal).

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Example 28: Expected Value for Poisson Distribution

Problem:
Find the expected number of events in a Poisson distribution with a rate of $\lambda =4$.

Answer:

Step 1: Given Data:
Rate $\lambda =4$.

Step 2: Solution:
For Poisson distribution, the expected value is:
$E(X)=\lambda =4$.

Step 3: Final Answer:
$E(X)=4$.

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Example 29: Variance of a Continuous Uniform Distribution

Problem:
For a continuous uniform distribution from 2 to 8, find the variance.

Answer:

Step 1: Given Data:
Lower bound $a=2$,
Upper bound $b=8$.

Step 2: Solution:
Using the formula for variance of a uniform distribution:
$Var(X)=\frac{(b–a{)}^2}{12}$

$Var(X)=\frac{(8–2{)}^2}{12}=\frac{36}{12}=3$.

Step 3: Final Answer:
$Var(X)=3$.

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Example 30: Probability from Exponential Distribution

Problem:
Find the probability that a random variable from an exponential distribution with $\lambda =0.2$ is less than 5.

Answer:

Step 1: Given Data:
Rate $\lambda =0.2$,
Value $X=5$.

Step 2: Solution:
Using the cumulative distribution function for exponential:
$P(X<x)=1–e^{-\lambda x}$

$P(X<5)=1–e^{-0.2\cdot 5}$.

Calculate:
$P(X<5)=1–e^{-1}\approx 1–0.3679=0.6321$.

Step 3: Final Answer:
$P(X<5)\approx 0.6321$.

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Example 31: Probability from Beta Distribution

Problem:
What is the probability of success in a Beta distribution with parameters $\alpha =2$ and $\beta =5$ at $x=0.5$?

Answer:

Step 1: Given Data:
Parameters $\alpha =2$,
$\beta =5$,
Value $x=0.5$.

Step 2: Solution:
Using the probability density function:
$f(x)=\frac{x^{\alpha –1}(1–x{)}^{\beta –1}}{B(\alpha ,\beta )}$,
where $B(\alpha ,\beta )=\frac{\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )}{\mathrm{\Gamma }(\alpha +\beta )}$.

Calculate:
$B(2,5)=\frac{\mathrm{\Gamma }(2)\mathrm{\Gamma }(5)}{\mathrm{\Gamma }(7)}=\frac{1\cdot 24}{720}=\frac{1}{30}$.

So:
$f(0.5)=\frac{(0.5{)}^{2–1}(1–0.5{)}^{5–1}}{B(2,5)}=\frac{(0.5{)}^1(0.5{)}^4}{\frac{1}{30}}=15\cdot {0.5}^5=15\cdot \frac{1}{32}=\frac{15}{32}\approx 0.46875$.

Step 3: Final Answer:
$f(0.5)\approx 0.46875$.

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Example 32: Mode of Binomial Distribution

Problem:
Find the mode of a binomial distribution with $n=10$ and $p=0.3$.

Answer:

Step 1: Given Data:
$n=10$,
$p=0.3$.

Step 2: Solution:
The mode of a binomial distribution is given by:
$k=\lfloor (n+1)p\rfloor =\lfloor (10+1)(0.3)\rfloor =\lfloor 3.3\rfloor =3$.

Step 3: Final Answer:
The mode is $3$.

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Example 33: Finding Probability with Joint Distribution

Problem:
If $X$ and $Y$ are independent random variables with $P(X=1)=0.6$ and $P(Y=2)=0.4$, find $P(X=1\cap Y=2)$.

Answer:

Step 1: Given Data:
$P(X=1)=0.6$,
$P(Y=2)=0.4$.

Step 2: Solution:
Since $X$ and $Y$ are independent:
$P(X=1\cap Y=2)=P(X=1)\cdot P(Y=2)$

$=0.6\cdot 0.4=0.24$.

Step 3: Final Answer:
$P(X=1\cap Y=2)=0.24$.

---

Example 34: CDF for Discrete Distribution

Problem:
For a discrete random variable $X$ with probabilities $P(1)=0.3$, $P(2)=0.5$, and $P(3)=0.2$, find the CDF.

Answer:

Step 1: Given Data:
Probabilities:
$P(1)=0.3$,
$P(2)=0.5$,
$P(3)=0.2$.

Step 2: Solution:
Calculate CDF:
$F(1)=P(X\le 1)=0.3$.

$F(2)=P(X\le 2)=0.3+0.5=0.8$.

$F(3)=P(X\le 3)=0.3+0.5+0.2=1.0$.

Step 3: Final Answer:
CDF values are:
$F(1)=0.3$,
$F(2)=0.8$,
$F(3)=1.0$.

---

Example 35: Probability of a Normal Variable Being Within Range

Problem:
Find the probability that a normally distributed random variable with $\mu =100$ and $\sigma =15$ falls between 90 and 110.

Answer:

Step 1: Given Data:
Mean $\mu =100$,
Standard deviation $\sigma =15$,
Range: $X_1=90$, $X_2=110$.

Step 2: Solution:
Calculate z-scores:
$Z_1=\frac{90–100}{15}=-0.67$,
$Z_2=\frac{110–100}{15}=0.67$.

Find probabilities:
$P(Z<0.67)\approx 0.7486$,
$P(Z<-0.67)\approx 0.2514$.

Calculate probability:
$P(90<X<110)=P(Z<0.67)–P(Z<-0.67)$

$=0.7486–0.2514=0.4972$.

Step 3: Final Answer:
$P(90<X<110)\approx 0.4972$.

---

Example 36: Probability in a Hypergeometric Distribution

Problem:
What is the probability of drawing 2 red balls from a box containing 5 red and 3 blue balls without replacement?

Answer:

Step 1: Given Data:
Total balls = 8,
Red balls = 5,
Blue balls = 3,
Draws = 2.

Step 2: Solution:
Using the hypergeometric formula:
$P(X=k)=\frac{(\genfrac{}{}{0ex}{}{K}{k})(\genfrac{}{}{0ex}{}{N-K}{n-k})}{(\genfrac{}{}{0ex}{}{N}{n})}$
$P(X=2)=\frac{(\genfrac{}{}{0ex}{}{5}{2})(\genfrac{}{}{0ex}{}{3}{0})}{(\genfrac{}{}{0ex}{}{8}{2})}$.

Calculate:
$(\genfrac{}{}{0ex}{}{5}{2})=10$,
$(\genfrac{}{}{0ex}{}{3}{0})=1$,
$(\genfrac{}{}{0ex}{}{8}{2})=28$.

So:
$P(X=2)=\frac{10\cdot 1}{28}=\frac{10}{28}=\frac{5}{14}$.

Step 3: Final Answer:
$P(X=2)=\frac{5}{14}$.

---

Example 37: Variance of Continuous Uniform Distribution

Problem:
Find the variance of a continuous uniform distribution between 0 and 6.

Answer:

Step 1: Given Data:
Lower bound $a=0$,
Upper bound $b=6$.

Step 2: Solution:
Using the formula for variance:
$Var(X)=\frac{(b–a{)}^2}{12}$

$Var(X)=\frac{(6–0{)}^2}{12}=\frac{36}{12}=3$.

Step 3: Final Answer:
$Var(X)=3$.

---

Example 38: Joint Probability of Two Events

Problem:
If $P(A)=0.5$, $P(B)=0.4$, and $P(A\cap B)=0.2$, find $P(A\cup B)$.

Answer:

Step 1: Given Data:
$P(A)=0.5$,
$P(B)=0.4$,
$P(A\cap B)=0.2$.

Step 2: Solution:
Using the formula:
$P(A\cup B)=P(A)+P(B)–P(A\cap B)$

$P(A\cup B)=0.5+0.4–0.2=0.7$.

Step 3: Final Answer:
$P(A\cup B)=0.7$.

---

Example 39: Probability Density Function of Normal Distribution

Problem:
What is the value of the PDF for a normal distribution with mean 0 and standard deviation 1 at $x=1$?

Answer:

Step 1: Given Data:
Mean $\mu =0$,
Standard deviation $\sigma =1$,
Value $x=1$.

Step 2: Solution:
Using the PDF formula:
$f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x–\mu {)}^2}{2{\sigma }^2}}$

Calculate:
$f(1)=\frac{1}{1\cdot \sqrt{2\pi }}{e}^{-\frac{(1–0{)}^2}{2\cdot 1^2}}$

$=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}}\approx 0.24197$.

Step 3: Final Answer:
$f(1)\approx 0.24197$.

---

Example 40: Finding the Median of a Normal Distribution

Problem:
What is the median of a normal distribution with $\mu =20$ and $\sigma =5$?

Answer:

Step 1: Given Data:
Mean $\mu =20$,
Standard deviation $\sigma =5$.

Step 2: Solution:
For a normal distribution, the median is equal to the mean.

Median = $\mu =20$.

Step 3: Final Answer:
Median = $20$.

---

Example 41: Exponential Distribution

Problem:
If the average time until an event is 3 minutes, what is the probability that the time until the next event is less than 2 minutes?

Answer:

Step 1: Given Data:
Rate $\lambda =\frac{1}{3}$,
Value $X=2$.

Step 2: Solution:
Using the cumulative distribution function:
$P(X<x)=1–e^{-\lambda x}$

$P(X<2)=1–e^{-\frac{2}{3}}$.

Calculate:
$P(X<2)\approx 1–0.5134\approx 0.4866$.

Step 3: Final Answer:
$P(X<2)\approx 0.4866$.

---

Example 42: Mode of a Continuous Distribution

Problem:
For a continuous uniform distribution between 1 and 5, find the mode.

Answer:

Step 1: Given Data:
Lower bound $a=1$,
Upper bound $b=5$.

Step 2: Solution:
In a continuous uniform distribution, every value in the interval is equally likely, so:

Mode = Any value in the interval, e.g., $1$ or $5$.

Step 3: Final Answer:
Mode = Any value in $[1,5]$.

---

Example 43: Geometric Distribution Probability

Problem:
What is the probability that the first success occurs on the 3rd trial with a success probability of 0.2?

Answer:

Step 1: Given Data:
$p=0.2$,
$k=3$.

Step 2: Solution:
Using the geometric formula:
$P(X=k)=(1–p{)}^{k-1}p$

$P(X=3)=(0.8{)}^2(0.2)=0.64\cdot 0.2=0.128$.

Step 3: Final Answer:
$P(X=3)=0.128$.

---

Example 44: Finding Variance for Exponential Distribution

Problem:
For an exponential distribution with $\lambda =0.5$, find the variance.

Answer:

Step 1: Given Data:
Rate $\lambda =0.5$.

Step 2: Solution:
Variance for exponential distribution is given by:
$Var(X)=\frac{1}{{\lambda }^2}$

$Var(X)=\frac{1}{(0.5{)}^2}=\frac{1}{0.25}=4$.

Step 3: Final Answer:
$Var(X)=4$.

---

Example 45: Probability of Success in Poisson Distribution

Problem:
If a restaurant gets an average of 2 customers per minute, what is the probability that exactly 1 customer arrives in the next minute?

Answer:

Step 1: Given Data:
$\lambda =2$,
$k=1$.

Step 2: Solution:
Using the Poisson formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

$P(X=1)=\frac{2^1{e}^{-2}}{1!}=2e^{-2}$.

Calculate:
$P(X=1)\approx 2\cdot 0.1353\approx 0.2706$.

Step 3: Final Answer:
$P(X=1)\approx 0.2706$.

---

Example 46: Normal Approximation to Binomial

Problem:
For a binomial distribution with $n=30$ and $p=0.2$, approximate the probability of getting at least 5 successes using the normal approximation.

Answer:

Step 1: Given Data:
$n=30$,
$p=0.2$.

Step 2: Solution:
Calculate mean and variance:
$\mu =np=30\cdot 0.2=6$
$Var(X)=np(1-p)=30\cdot 0.2\cdot 0.8=4.8$.

Calculate standard deviation:
$\sigma =\sqrt{4.8}\approx 2.19$.

Use continuity correction for normal approximation:
Find $P(X\ge 5)$:
$P(X\ge 5)\approx P(Z\ge \frac{4.5–6}{2.19})\approx P(Z\ge -0.68)$.

Use z-table:
$P(Z\ge -0.68)\approx 0.7517$.

Step 3: Final Answer:
$P(X\ge 5)\approx 0.7517$.

---

Example 47: Conditional Probability

Problem:
In a bag of 5 red and 3 blue marbles, what is the probability of drawing a blue marble given that a red marble was drawn first?

Answer:

Step 1: Given Data:
Total marbles = 8,
Red marbles = 5,
Blue marbles = 3.

Step 2: Solution:
After drawing a red marble, there are 7 marbles left:
Red marbles = 4,
Blue marbles = 3.

Using conditional probability:
$P(\text{Blue}|\text{Red})=\frac{P(\text{Blue}\cap \text{Red})}{P(\text{Red})}$

$P(\text{Blue}|\text{Red})=\frac{3}{7}$.

Step 3: Final Answer:
$P(\text{Blue}|\text{Red})=\frac{3}{7}$.

---

Example 48: Joint Probability with Conditional

Problem:
What is the probability of rolling a 2 on a die and then drawing a red card from a standard deck of cards?

Answer:

Step 1: Given Data:
$P(\text{Rolling a 2})=\frac{1}{6}$,
$P(\text{Drawing a Red Card})=\frac{26}{52}=\frac{1}{2}$.

Step 2: Solution:
Since the events are independent:
$P(\text{Rolling a 2 and Drawing Red})=P(\text{Rolling a 2})\cdot P(\text{Drawing Red})$

$=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$.

Step 3: Final Answer:
$P(\text{Rolling a 2 and Drawing Red})=\frac{1}{12}$.

---

Example 49: Exponential Distribution Probability

Problem:
What is the probability that a customer waits less than 4 minutes at a service desk, given that the average waiting time is 3 minutes?

Answer:

Step 1: Given Data:
Rate $\lambda =\frac{1}{3}$,
Value $X=4$.

Step 2: Solution:
Using the CDF for exponential distribution:
$P(X<x)=1–e^{-\lambda x}$

$P(X<4)=1–e^{-\frac{1}{3}\cdot 4}=1–e^{-\frac{4}{3}}$.

Calculate:
$P(X<4)\approx 1–0.2636=0.7364$.

Step 3: Final Answer:
$P(X<4)\approx 0.7364$.

---

Example 50: Variance of Normal Distribution

Problem:
Find the variance of a normal distribution with mean 25 and standard deviation 6.

Answer:

Step 1: Given Data:
Mean $\mu =25$,
Standard deviation $\sigma =6$.

Step 2: Solution:
Calculate variance:
$Var(X)={\sigma }^2=6^2=36$.

Step 3: Final Answer:
$Var(X)=36$.

---

Example 51: Probability of Exceeding a Value

Problem:
If the average lifespan of a battery is 200 hours with a standard deviation of 30 hours, what is the probability that a randomly selected battery lasts more than 250 hours?

Answer:

Step 1: Given Data:
Mean $\mu =200$,
Standard deviation $\sigma =30$,
Value $X=250$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{250–200}{30}=\frac{50}{30}\approx 1.67$.

Use z-tables to find:
$P(Z>1.67)\approx 1–0.9525=0.0475$.

Step 3: Final Answer:
$P(X>250)\approx 0.0475$.

---

Example 52: Probability of Rolling a Die

Problem:
What is the probability of rolling an even number on a six-sided die?

Answer:

Step 1: Given Data:
Total outcomes = 6,
Favorable outcomes = 3 (2, 4, 6).

Step 2: Solution:
Using the probability formula:
$P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{Even})=\frac{3}{6}=\frac{1}{2}$.

Step 3: Final Answer:
$P(\text{Even})=\frac{1}{2}$.

---

Example 53: Binomial Distribution with More Trials

Problem:
If you flip a coin 8 times, what is the probability of getting exactly 6 heads?

Answer:

Step 1: Given Data:
$n=8$,
$k=6$,
$p=0.5$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=6)=(\genfrac{}{}{0ex}{}{8}{6})(0.5{)}^6(0.5{)}^2$.

Calculate:
$(\genfrac{}{}{0ex}{}{8}{6})=28$.

So:
$P(X=6)=28\cdot (0.5{)}^8=28\cdot \frac{1}{256}=\frac{28}{256}=\frac{7}{64}$.

Step 3: Final Answer:
$P(X=6)=\frac{7}{64}$.

---

Example 54: Conditional Probability with Replacement

Problem:
In a deck of cards, what is the probability of drawing an Ace given that the first card drawn was not an Ace?

Answer:

Step 1: Given Data:
Total cards = 52,
Non-Aces = 48,
Aces = 4.

Step 2: Solution:
After drawing a non-Ace, there are still 52 cards:
$P(Ace|NotAce)=\frac{P(Ace)}{P(NotAce)}=\frac{4}{48}=\frac{1}{12}$.

Step 3: Final Answer:
$P(Ace|NotAce)=\frac{1}{12}$.

---

Example 55: Variance of a Poisson Distribution

Problem:
Find the variance of a Poisson distribution where $\lambda =3$.

Answer:

Step 1: Given Data:
$\lambda =3$.

Step 2: Solution:
The variance of a Poisson distribution is given by:
$Var(X)=\lambda =3$.

Step 3: Final Answer:
$Var(X)=3$.

---

Example 56: Probability of a Normal Variable Falling Below a Value

Problem:
What is the probability that a normally distributed random variable with $\mu =50$ and $\sigma =10$ is less than 40?

Answer:

Step 1: Given Data:
Mean $\mu =50$,
Standard deviation $\sigma =10$,
Value $X=40$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{40–50}{10}=-1$.

Use z-tables to find:
$P(Z<-1)\approx 0.1587$.

Step 3: Final Answer:
$P(X<40)\approx 0.1587$.

---

Example 57: Probability of Getting a Specific Number of Events

Problem:
If a restaurant serves an average of 8 meals per hour, what is the probability that it serves exactly 5 meals in an hour?

Answer:

Step 1: Given Data:
$\lambda =8$,
$k=5$.

Step 2: Solution:
Using the Poisson formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

$P(X=5)=\frac{8^5{e}^{-8}}{5!}$.

Calculate:
$P(X=5)=\frac{32768e^{-8}}{120}$.
Using $e^{-8}\approx 0.00033546$:
$P(X=5)\approx \frac{32768\cdot 0.00033546}{120}\approx 0.914$.

Step 3: Final Answer:
$P(X=5)\approx 0.914$.

---

Example 58: Finding Mode of a Probability Distribution

Problem:
For the discrete distribution $P(1)=0.1$, $P(2)=0.5$, $P(3)=0.4$, find the mode.

Answer:

Step 1: Given Data:
Probabilities:
$P(1)=0.1$,
$P(2)=0.5$,
$P(3)=0.4$.

Step 2: Solution:
Identify the value with the highest probability:
The mode is $2$.

Step 3: Final Answer:
The mode is $2$.

---

Example 59: Finding the Mean of a Binomial Distribution

Problem:
For a binomial distribution with $n=15$ and $p=0.3$, what is the expected value?

Answer:

Step 1: Given Data:
$n=15$,
$p=0.3$.

Step 2: Solution:
Using the formula for expected value:
$E(X)=n\cdot p$

$E(X)=15\cdot 0.3=4.5$.

Step 3: Final Answer:
$E(X)=4.5$.

---

Example 60: Finding Variance in Binomial Distribution

Problem:
Find the variance of a binomial distribution with $n=20$ and $p=0.6$.

Answer:

Step 1: Given Data:
$n=20$,
$p=0.6$.

Step 2: Solution:
Using the variance formula:
$Var(X)=n\cdot p\cdot (1–p)$

$Var(X)=20\cdot 0.6\cdot (1–0.6)=20\cdot 0.6\cdot 0.4=4.8$.

Step 3: Final Answer:
$Var(X)=4.8$.

---

Example 61: Cumulative Probability from a Distribution

Problem:
Find the cumulative probability for a normal distribution at $x=100$ with $\mu =90$ and $\sigma =10$.

Answer:

Step 1: Given Data:
Mean $\mu =90$,
Standard deviation $\sigma =10$,
Value $X=100$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{100–90}{10}=1$.

Use z-tables to find:
$P(Z<1)\approx 0.8413$.

Step 3: Final Answer:
$P(X<100)\approx 0.8413$.

---

Example 62: Probability of a Uniform Distribution

Problem:
If a random variable is uniformly distributed from 3 to 7, what is the probability it is less than 4?

Answer:

Step 1: Given Data:
Lower bound $a=3$,
Upper bound $b=7$.

Step 2: Solution:
Total range:
$Range=b–a=7–3=4$.

Specific range:
$Rang{e}_{<4}=4–3=1$.

Calculate probability:
$P(X<4)=\frac{Rang{e}_{<4}}{Range}=\frac{1}{4}=0.25$.

Step 3: Final Answer:
$P(X<4)=0.25$.

---

Example 63: Mode of a Poisson Distribution

Problem:
What is the mode of a Poisson distribution with $\lambda =2$?

Answer:

Step 1: Given Data:
$\lambda =2$.

Step 2: Solution:
The mode of a Poisson distribution is given by:
$mode=\lfloor \lambda \rfloor$

If $\lambda$ is an integer, the mode is $\lambda$ or $\lambda –1$.

So:
$mode=\lfloor 2\rfloor =2$.

Step 3: Final Answer:
The mode is $2$.

---

Example 64: Expected Value of a Discrete Random Variable

Problem:
For the random variable $X$ with values 1, 4, 6 and probabilities $0.2$, $0.5$, $0.3$, find the expected value.

Answer:

Step 1: Given Data:
Values: $1,4,6$
Probabilities: $0.2,0.5,0.3$.

Step 2: Solution:
Calculate expected value:
$E(X)=(1\cdot 0.2)+(4\cdot 0.5)+(6\cdot 0.3)$

$E(X)=0.2+2+1.8=4$.

Step 3: Final Answer:
$E(X)=4$.

---

Example 65: Variance of a Discrete Random Variable

Problem:
Find the variance for a discrete random variable with values $1,2,3$ and probabilities $0.1$, $0.3$, $0.6$.

Answer:

Step 1: Given Data:
Values: $1,2,3$
Probabilities: $0.1,0.3,0.6$.

Step 2: Solution:
Calculate expected value:
$E(X)=(1\cdot 0.1)+(2\cdot 0.3)+(3\cdot 0.6)$

$E(X)=0.1+0.6+1.8=2.5$.

Calculate $E(X^2)$:
$E(X^2)=(1^2\cdot 0.1)+(2^2\cdot 0.3)+(3^2\cdot 0.6)$

$E(X^2)=0.1+1.2+5.4=6.7$.

Calculate variance:
$Var(X)=E(X^2)–(E(X){)}^2=6.7–(2.5{)}^2=6.7–6.25=0.45$.

Step 3: Final Answer:
$Var(X)=0.45$.

---

Example 66: Probability of a Normal Variable Being Greater Than a Value

Problem:
What is the probability that a normally distributed random variable with $\mu =70$ and $\sigma =10$ is greater than 80?

Answer:

Step 1: Given Data:
Mean $\mu =70$,
Standard deviation $\sigma =10$,
Value $X=80$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{80–70}{10}=1$.

Use z-tables to find:
$P(Z>1)\approx 1–0.8413=0.1587$.

Step 3: Final Answer:
$P(X>80)\approx 0.1587$.

---

Example 67: Finding Probability Using Central Limit Theorem

Problem:
If a sample of 50 has a mean of 100 and standard deviation of 15, what is the probability that the sample mean is less than 98?

Answer:

Step 1: Given Data:
Sample size $n=50$,
Mean $\mu =100$,
Standard deviation $\sigma =15$.

Step 2: Solution:
Calculate standard error:
$SE=\frac{\sigma }{\sqrt{n}}=\frac{15}{\sqrt{50}}\approx 2.1213$.

Calculate z-score:
$Z=\frac{98–100}{SE}=\frac{-2}{2.1213}\approx -0.94$.

Use z-tables to find:
$P(Z<-0.94)\approx 0.1736$.

Step 3: Final Answer:
$P(\text{Sample mean}<98)\approx 0.1736$.

---

Example 68: Probability of a Single Event

Problem:
What is the probability of drawing an Ace from a standard deck of 52 cards?

Answer:

Step 1: Given Data:
Total cards = 52,
Aces = 4.

Step 2: Solution:
Using the probability formula:
$P(Ace)=\frac{\text{Number of Aces}}{\text{Total cards}}$

$P(Ace)=\frac{4}{52}=\frac{1}{13}$.

Step 3: Final Answer:
$P(Ace)=\frac{1}{13}$.

---

Example 69: Expected Value from Joint Distribution

Problem:
If $X$ and $Y$ are independent random variables where $E(X)=3$ and $E(Y)=5$, what is $E(X+Y)$?

Answer:

Step 1: Given Data:
$E(X)=3$,
$E(Y)=5$.

Step 2: Solution:
Using the linearity of expectation:
$E(X+Y)=E(X)+E(Y)$

$E(X+Y)=3+5=8$.

Step 3: Final Answer:
$E(X+Y)=8$.

---

Example 70: Probability of a Bimodal Distribution

Problem:
For the discrete random variable $X$ with $P(1)=0.2$, $P(2)=0.2$, $P(3)=0.5$, find the mode.

Answer:

Step 1: Given Data:
Probabilities:
$P(1)=0.2$,
$P(2)=0.2$,
$P(3)=0.5$.

Step 2: Solution:
Identify the value with the highest probability:
The mode is $3$.

Step 3: Final Answer:
The mode is $3$.

---

Example 71: Variance of Normal Distribution

Problem:
If a normal distribution has $\mu =15$ and $\sigma =4$, find the variance.

Answer:

Step 1: Given Data:
Mean $\mu =15$,
Standard deviation $\sigma =4$.

Step 2: Solution:
Calculate variance:
$Var(X)={\sigma }^2=4^2=16$.

Step 3: Final Answer:
$Var(X)=16$.

---

Example 72: Poisson Distribution for More Events

Problem:
If a store has an average of 6 customers per hour, what is the probability that exactly 4 customers arrive in the next hour?

Answer:

Step 1: Given Data:
$\lambda =6$,
$k=4$.

Step 2: Solution:
Using the Poisson formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

$P(X=4)=\frac{6^4{e}^{-6}}{4!}$.

Calculate:
$P(X=4)=\frac{1296e^{-6}}{24}$.
Using $e^{-6}\approx 0.002478752$:
$P(X=4)\approx \frac{1296\cdot 0.002478752}{24}\approx 0.128$.

Step 3: Final Answer:
$P(X=4)\approx 0.128$.

---

Example 73: Probability in a Geometric Distribution

Problem:
What is the probability that the first success occurs on the 5th trial with a success probability of 0.1?

Answer:

Step 1: Given Data:
$p=0.1$,
$k=5$.

Step 2: Solution:
Using the geometric distribution formula:
$P(X=k)=(1–p{)}^{k-1}p$

$P(X=5)=(0.9{)}^4(0.1)$.

Calculate:
$P(X=5)=0.6561\cdot 0.1=0.06561$.

Step 3: Final Answer:
$P(X=5)\approx 0.06561$.

---

Example 74: Probability of Binomial Distribution

Problem:
If a die is rolled 4 times, what is the probability of getting exactly 2 fours?

Answer:

Step 1: Given Data:
$n=4$,
$k=2$,
$p=\frac{1}{6}$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=2)=(\genfrac{}{}{0ex}{}{4}{2}){\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^2$.

Calculate:
$(\genfrac{}{}{0ex}{}{4}{2})=6$.

So:
$P(X=2)=6\cdot \frac{1}{36}\cdot \frac{25}{36}=6\cdot \frac{25}{1296}=\frac{150}{1296}=\frac{25}{216}$.

Step 3: Final Answer:
$P(X=2)=\frac{25}{216}$.

---

Example 75: Probability of Exceeding a Certain Value

Problem:
For a normal distribution with $\mu =30$ and $\sigma =5$, what is the probability of scoring above 35?

Answer:

Step 1: Given Data:
Mean $\mu =30$,
Standard deviation $\sigma =5$,
Value $X=35$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{X–\mu }{\sigma }=\frac{35–30}{5}=1$.

Use z-tables to find:
$P(Z>1)\approx 1–P(Z<1)\approx 1–0.8413=0.1587$.

Step 3: Final Answer:
$P(X>35)\approx 0.1587$.

---

Example 76: Probability of a Continuous Uniform Distribution

Problem:
If a continuous random variable is uniformly distributed between 2 and 8, what is the probability of it being between 4 and 6?

Answer:

Step 1: Given Data:
Lower bound $a=2$,
Upper bound $b=8$.

Step 2: Solution:
Total range:
$Range=b–a=8–2=6$.

Specific range:
$Rang{e}_{4-6}=6–4=2$.

Calculate probability:
$P(4<X<6)=\frac{Rang{e}_{4-6}}{Range}=\frac{2}{6}=\frac{1}{3}$.

Step 3: Final Answer:
$P(4<X<6)=\frac{1}{3}$.

---

Example 77: Probability from a Beta Distribution

Problem:
What is the probability density function value for a Beta distribution with parameters $\alpha =2$, $\beta =5$ at $x=0.3$?

Answer:

Step 1: Given Data:
Parameters $\alpha =2$,
$\beta =5$,
Value $x=0.3$.

Step 2: Solution:
Using the PDF formula:
$f(x)=\frac{x^{\alpha –1}(1–x{)}^{\beta –1}}{B(\alpha ,\beta )}$

$B(2,5)=\frac{\mathrm{\Gamma }(2)\mathrm{\Gamma }(5)}{\mathrm{\Gamma }(7)}=\frac{1\cdot 24}{720}=\frac{1}{30}$.

Calculate:
$f(0.3)=\frac{(0.3{)}^1(0.7{)}^4}{\frac{1}{30}}$.

Calculate:
$=30\cdot (0.3)\cdot (0.7{)}^4\approx 30\cdot 0.3\cdot 0.2401=1.441$.

Step 3: Final Answer:
$f(0.3)\approx 1.441$.

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Example 78: Finding Variance of a Hypergeometric Distribution

Problem:
For a hypergeometric distribution with $N=20$, $K=10$, and $n=5$, find the variance.

Answer:

Step 1: Given Data:
Total population $N=20$,
Number of successes $K=10$,
Sample size $n=5$.

Step 2: Solution:
Using the formula for variance:
$Var(X)=n\cdot \frac{K}{N}\cdot \frac{N–K}{N}\cdot \frac{N–n}{N–1}$

$Var(X)=5\cdot \frac{10}{20}\cdot \frac{10}{20}\cdot \frac{15}{19}$.

Calculate:
$Var(X)=5\cdot 0.5\cdot 0.5\cdot \frac{15}{19}=\frac{5\cdot 0.25\cdot 15}{19}\approx 0.9868$.

Step 3: Final Answer:
$Var(X)\approx 0.9868$.

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Example 79: Probability of Getting More Than a Certain Number of Successes

Problem:
If a coin is flipped 10 times, what is the probability of getting more than 8 heads?

Answer:

Step 1: Given Data:
$n=10$,
$k>8$.

Step 2: Solution:
Calculate the probabilities for $k=9$ and $k=10$:
$P(X=9)=(\genfrac{}{}{0ex}{}{10}{9})(0.5{)}^9(0.5{)}^1=10\cdot \frac{1}{1024}=\frac{10}{1024}$.

$P(X=10)=(\genfrac{}{}{0ex}{}{10}{10})(0.5{)}^{10}=1\cdot \frac{1}{1024}=\frac{1}{1024}$.

Now, calculate total probability:
$P(X>8)=P(X=9)+P(X=10)$
$=\frac{10}{1024}+\frac{1}{1024}=\frac{11}{1024}\approx 0.01074$.

Step 3: Final Answer:
$P(X>8)\approx 0.01074$.

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Example 80: Finding Probability from the CDF of a Normal Distribution

Problem:
For a normal distribution with $\mu =40$ and $\sigma =8$, find the probability that a value is less than 36.

Answer:

Step 1: Given Data:
Mean $\mu =40$,
Standard deviation $\sigma =8$,
Value $X=36$.

Step 2: Solution:
Calculate z-score:
$Z=\frac{36–40}{8}=-0.5$.

Use z-tables to find:
$P(Z<-0.5)\approx 0.3085$.

Step 3: Final Answer:
$P(X<36)\approx 0.3085$.

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Example 81: Variance in Exponential Distribution

Problem:
What is the variance of an exponential distribution with $\lambda =0.25$?

Answer:

Step 1: Given Data:
Rate $\lambda =0.25$.

Step 2: Solution:
The variance of an exponential distribution is given by:
$Var(X)=\frac{1}{{\lambda }^2}$

$Var(X)=\frac{1}{(0.25{)}^2}=\frac{1}{0.0625}=16$.

Step 3: Final Answer:
$Var(X)=16$.

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Example 82: Probability of an Event in a Binomial Distribution

Problem:
In 10 flips of a fair coin, what is the probability of getting at least 7 heads?

Answer:

Step 1: Given Data:
$n=10$,
$k\ge 7$.

Step 2: Solution:
Calculate probabilities for $k=7,8,9,10$.
$P(X=7)=(\genfrac{}{}{0ex}{}{10}{7})(0.5{)}^{10}=\frac{120}{1024}$.

$P(X=8)=(\genfrac{}{}{0ex}{}{10}{8})(0.5{)}^{10}=\frac{45}{1024}$.

$P(X=9)=(\genfrac{}{}{0ex}{}{10}{9})(0.5{)}^{10}=\frac{10}{1024}$.

$P(X=10)=(\genfrac{}{}{0ex}{}{10}{10})(0.5{)}^{10}=\frac{1}{1024}$.

Now add:
$P(X\ge 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)$
$=\frac{120+45+10+1}{1024}=\frac{176}{1024}\approx 0.1719$.

Step 3: Final Answer:
$P(X\ge 7)\approx 0.1719$.

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Example 83: Conditional Probability in a Deck of Cards

Problem:
In a deck of cards, what is the probability of drawing a heart given that the card drawn is a red card?

Answer:

Step 1: Given Data:
Total red cards = 26 (13 hearts, 13 diamonds).
Favorable outcomes = 13 (hearts).

Step 2: Solution:
Using conditional probability:
$P(\text{Heart}|\text{Red})=\frac{P(\text{Heart}\cap \text{Red})}{P(\text{Red})}$

$P(\text{Heart}|\text{Red})=\frac{13/52}{26/52}=\frac{13}{26}=\frac{1}{2}$.

Step 3: Final Answer:
$P(\text{Heart}|\text{Red})=\frac{1}{2}$.

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Example 84: Probability in Joint Distribution

Problem:
What is the probability of drawing a King and a Queen in a 52-card deck?

Answer:

Step 1: Given Data:
Total cards = 52,
Kings = 4,
Queens = 4.

Step 2: Solution:
$P(\text{King})=\frac{4}{52}$,
$P(\text{Queen})=\frac{4}{52}$.

Since drawing without replacement:
$P(\text{King and then Queen})=P(\text{King})\cdot P(\text{Queen given King})$
$=\frac{4}{52}\cdot \frac{4}{51}$.

Calculate:
$P(\text{King and Queen})=\frac{4\cdot 4}{52\cdot 51}=\frac{16}{2652}=\frac{4}{663}$.

Step 3: Final Answer:
$P(\text{King and Queen})=\frac{4}{663}$.

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Example 85: Finding Probability of an Event

Problem:
If a bag contains 5 red balls and 7 blue balls, what is the probability of drawing a red ball?

Answer:

Step 1: Given Data:
Total balls = 12,
Red balls = 5.

Step 2: Solution:
Using the probability formula:
$P(Red)=\frac{\text{Number of Red Balls}}{\text{Total Balls}}$

$P(Red)=\frac{5}{12}$.

Step 3: Final Answer:
$P(Red)=\frac{5}{12}$.

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Example 86: Probability of a Discrete Random Variable

Problem:
What is the probability of getting 3 successes in 5 trials with a success probability of 0.4?

Answer:

Step 1: Given Data:
$n=5$,
$k=3$,
$p=0.4$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=3)=(\genfrac{}{}{0ex}{}{5}{3})(0.4{)}^3(0.6{)}^2$.

Calculate:
$(\genfrac{}{}{0ex}{}{5}{3})=10$.

So:
$P(X=3)=10\cdot (0.064)\cdot (0.36)=10\cdot 0.02304\approx 0.2304$.

Step 3: Final Answer:
$P(X=3)\approx 0.2304$.

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Example 87: Finding Expected Value in a Distribution

Problem:
For a discrete random variable with values 1, 2, 3 and probabilities 0.1, 0.6, 0.3, find the expected value.

Answer:

Step 1: Given Data:
Values: $1,2,3$
Probabilities: $0.1,0.6,0.3$.

Step 2: Solution:
Calculate expected value:
$E(X)=(1\cdot 0.1)+(2\cdot 0.6)+(3\cdot 0.3)$

$E(X)=0.1+1.2+0.9=2.2$.

Step 3: Final Answer:
$E(X)=2.2$.

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Example 88: Cumulative Distribution Function of a Discrete Variable

Problem:
For a random variable $X$ with probabilities $P(1)=0.2$, $P(2)=0.5$, and $P(3)=0.3$, find the CDF.

Answer:

Step 1: Given Data:
Probabilities:
$P(1)=0.2$,
$P(2)=0.5$,
$P(3)=0.3$.

Step 2: Solution:
Calculate CDF:
$F(1)=P(X\le 1)=0.2$.

$F(2)=P(X\le 2)=0.2+0.5=0.7$.

$F(3)=P(X\le 3)=0.2+0.5+0.3=1$.

Step 3: Final Answer:
CDF values are:
$F(1)=0.2$,
$F(2)=0.7$,
$F(3)=1$.

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Example 89: Probability from a CDF of a Continuous Distribution

Problem:
For a continuous distribution with CDF $F(x)=0.5$ at $x=10$, what is the probability that $X$ is less than 10?

Answer:

Step 1: Given Data:
$F(10)=0.5$.

Step 2: Solution:
$P(X<10)=F(10)$.

Step 3: Final Answer:
$P(X<10)=0.5$.

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Example 90: Probability of Rolling a Die

Problem:
What is the probability of rolling a number greater than 4 on a six-sided die?

Answer:

Step 1: Given Data:
Total outcomes = 6,
Favorable outcomes = 2 (5, 6).

Step 2: Solution:
Using the probability formula:
$P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{> 4})=\frac{2}{6}=\frac{1}{3}$.

Step 3: Final Answer:
$P(\text{> 4})=\frac{1}{3}$.

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Example 91: Variance of a Discrete Distribution

Problem:
Find the variance of a discrete random variable with values $2,4,6$ and probabilities $0.3,0.4,0.3$.

Answer:

Step 1: Given Data:
Values: $2,4,6$
Probabilities: $0.3,0.4,0.3$.

Step 2: Solution:
Calculate expected value:
$E(X)=(2\cdot 0.3)+(4\cdot 0.4)+(6\cdot 0.3)$

$E(X)=0.6+1.6+1.8=4$.

Calculate $E(X^2)$:
$E(X^2)=(2^2\cdot 0.3)+(4^2\cdot 0.4)+(6^2\cdot 0.3)$

$E(X^2)=(4\cdot 0.3)+(16\cdot 0.4)+(36\cdot 0.3)$

$=1.2+6.4+10.8=18.4$.

Calculate variance:
$Var(X)=E(X^2)–(E(X){)}^2=18.4–4^2=18.4–16=2.4$.

Step 3: Final Answer:
$Var(X)=2.4$.

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Example 92: Finding Probability of Poisson Distribution

Problem:
A factory has an average of 5 machine breakdowns per month. What is the probability of exactly 2 breakdowns in a month?

Answer:

Step 1: Given Data:
$\lambda =5$,
$k=2$.

Step 2: Solution:
Using the Poisson formula:
$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

$P(X=2)=\frac{5^2{e}^{-5}}{2!}=\frac{25e^{-5}}{2}$.

Calculate:
$=\frac{25\cdot 0.006737947}{2}\approx 0.0842$.

Step 3: Final Answer:
$P(X=2)\approx 0.0842$.

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Example 93: Probability of a Success in a Binomial Distribution

Problem:
If you flip a coin 10 times, what is the probability of getting exactly 1 head?

Answer:

Step 1: Given Data:
$n=10$,
$k=1$,
$p=0.5$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=1)=(\genfrac{}{}{0ex}{}{10}{1})(0.5{)}^1(0.5{)}^9$.

Calculate:
$(\genfrac{}{}{0ex}{}{10}{1})=10$.

So:
$P(X=1)=10\cdot (0.5{)}^{10}=10\cdot \frac{1}{1024}=\frac{10}{1024}=\frac{5}{512}$.

Step 3: Final Answer:
$P(X=1)=\frac{5}{512}$.

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Example 94: Finding the Mean of a Distribution

Problem:
For a random variable with values $0,1,2$ and probabilities $0.2,0.5,0.3$, find the mean.

Answer:

Step 1: Given Data:
Values: $0,1,2$
Probabilities: $0.2,0.5,0.3$.

Step 2: Solution:
Calculate mean:
$E(X)=(0\cdot 0.2)+(1\cdot 0.5)+(2\cdot 0.3)$

$E(X)=0+0.5+0.6=1.1$.

Step 3: Final Answer:
$E(X)=1.1$.

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Example 95: Probability of an Event in a Continuous Uniform Distribution

Problem:
If a continuous random variable is uniformly distributed between 5 and 15, what is the probability of it being less than 10?

Answer:

Step 1: Given Data:
Lower bound $a=5$,
Upper bound $b=15$.

Step 2: Solution:
Total range:
$Range=b–a=15–5=10$.

Specific range:
$Rang{e}_{<10}=10–5=5$.

Calculate probability:
$P(X<10)=\frac{Rang{e}_{<10}}{Range}=\frac{5}{10}=0.5$.

Step 3: Final Answer:
$P(X<10)=0.5$.

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Example 96: Mode of a Binomial Distribution

Problem:
Find the mode of a binomial distribution with $n=12$ and $p=0.5$.

Answer:

Step 1: Given Data:
$n=12$,
$p=0.5$.

Step 2: Solution:
Calculate the mode using:
$k=\lfloor (n+1)p\rfloor$

$k=\lfloor (12+1)(0.5)\rfloor =\lfloor 6.5\rfloor =6$.

Step 3: Final Answer:
The mode is $6$.

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Example 97: Finding Probability in Exponential Distribution

Problem:
If the average time between arrivals at a service desk is 8 minutes, what is the probability that the next arrival is in less than 5 minutes?

Answer:

Step 1: Given Data:
Rate $\lambda =\frac{1}{8}$,
Value $X=5$.

Step 2: Solution:
Using the cumulative distribution function:
$P(X<x)=1–e^{-\lambda x}$

$P(X<5)=1–e^{-\frac{5}{8}}$.

Calculate:
$P(X<5)\approx 1–e^{-0.625}\approx 1–0.5353=0.4647$.

Step 3: Final Answer:
$P(X<5)\approx 0.4647$.

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Example 98: Probability of Independent Events

Problem:
What is the probability of flipping a coin and rolling a die simultaneously, resulting in heads and a 3?

Answer:

Step 1: Given Data:
$P(\text{Heads})=\frac{1}{2}$,
$P(\text{Rolling a 3})=\frac{1}{6}$.

Step 2: Solution:
Since the events are independent:
$P(\text{Heads and 3})=P(\text{Heads})\cdot P(\text{Rolling a 3})$

$=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$.

Step 3: Final Answer:
$P(\text{Heads and 3})=\frac{1}{12}$.

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Example 99: Finding Variance of a Normal Distribution

Problem:
If a normal distribution has $\mu =60$ and $\sigma =10$, find the variance.

Answer:

Step 1: Given Data:
Mean $\mu =60$,
Standard deviation $\sigma =10$.

Step 2: Solution:
Calculate variance:
$Var(X)={\sigma }^2={10}^2=100$.

Step 3: Final Answer:
$Var(X)=100$.

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Example 100: Probability of a Binomial Event

Problem:
In a binomial experiment with $n=5$ and $p=0.7$, what is the probability of getting exactly 4 successes?

Answer:

Step 1: Given Data:
$n=5$,
$k=4$,
$p=0.7$.

Step 2: Solution:
Using the binomial formula:
$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

$P(X=4)=(\genfrac{}{}{0ex}{}{5}{4})(0.7{)}^4(0.3{)}^1$.

Calculate:
$(\genfrac{}{}{0ex}{}{5}{4})=5$.

So:
$P(X=4)=5\cdot (0.7{)}^4\cdot (0.3)=5\cdot 0.2401\cdot 0.3\approx 0.36015$.

Step 3: Final Answer:
$P(X=4)\approx 0.36015$.