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.
Difference Between Discrete and Continuous Distributions
- 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.
- 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.
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
Discrete Probability Distributions
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.
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) = \binom{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) = \binom{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 $
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
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{125 e^{-5}}{6} $
Step 3: Final Answer:
$ P(X = 3) \approx 0.1404 $
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{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} $
Multinomial Distribution
The multinomial distribution generalizes the binomial distribution by allowing more than two possible outcomes.
Continuous Probability Distributions
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.
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 $
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 \leq x \leq b $
Exponential Distribution
The exponential distribution models the time between events in a Poisson process.
Formula:
$ f(x) = \lambda e^{-\lambda x} \text{ for } x \geq 0 $
Gamma Distribution
The gamma distribution generalizes the exponential distribution for the time to multiple events.
Beta Distribution
The beta distribution is used to model random variables that are bounded between 0 and 1.
Weibull Distribution
The Weibull distribution is used in reliability analysis and survival studies.
Probability Density Function (PDF)
The PDF gives the likelihood of a continuous random variable taking on a specific value.
Cumulative Distribution Function (CDF)
The CDF gives the probability that a random variable is less than or equal to a particular value.
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}) $
Expected Value and Variance for Discrete and Continuous Distributions
- Expected Value (Discrete):
$ E(X) = \sum x_i P(x_i) $ - Variance (Discrete):
$ \text{Var}(X) = E(X^2) – (E(X))^2 $ - Expected Value (Continuous):
$ E(X) = \int_{-\infty}^{\infty} x f(x) , dx $ - Variance (Continuous):
$ \text{Var}(X) = \int_{-\infty}^{\infty} (x – \mu)^2 f(x) , dx $
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.
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.
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.
FAQs
- 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.
- 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).
- 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.
- 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_{-\infty}^{\infty} x f(x) , dx $
- The expected value for a continuous distribution is calculated using the integral:
- 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.
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:
$P(4) = \frac{1}{6}$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 5) = \binom{10}{5} (0.5)^5 (0.5)^{5}$.
Calculate:
$\binom{10}{5} = 252$.
So:
$P(X = 5) = 252 \cdot \left(\frac{1}{2}\right)^{10} = 252 \cdot \frac{1}{1024} \approx 0.2461$.
Step 3: Final Answer:
$P(X = 5) \approx 0.2461$.
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:
$= \frac{81 e^{-3}}{24} \approx \frac{81 \cdot 0.0498}{24} \approx 0.1680$.
Step 3: Final Answer:
$P(X = 4) \approx 0.1680$.
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:
$P(X < 70) \approx 0.3085$.
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:
$Range_{3-7} = 7 – 3 = 4$.
Now calculate probability:
$P(3 < X < 7) = \frac{Range_{3-7}}{Range} = \frac{4}{8} = 0.5$.
Step 3: Final Answer:
$P(3 < X < 7) = 0.5$.
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:
$E(X) = 2.1$.
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:
$Var(X) = 0.44$.
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:
$F(1) = P(X \leq 1) = 0.2$.
$F(2) = P(X \leq 2) = 0.2 + 0.5 = 0.7$.
$F(3) = P(X \leq 3) = 0.2 + 0.5 + 0.3 = 1$.
Step 3: Final Answer:
CDF values are:
$F(1) = 0.2$,
$F(2) = 0.7$,
$F(3) = 1$.
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:
$P(\text{Tea or Coffee}) = 0.6$.
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:
$P(X > 120) \approx 0.0918$.
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:
$P(X = 4) \approx 0.1029$.
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:
$Range_{12-15} = 15 – 12 = 3$.
Now calculate probability:
$P(12 < X < 15) = \frac{Range_{12-15}}{Range} = \frac{3}{10} = 0.3$.
Step 3: Final Answer:
$P(12 < X < 15) = 0.3$.
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:
$E(X) = 10$.
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:
$Var(X) = 25$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 2) = \binom{6}{2} (0.5)^2 (0.5)^{4}$.
Calculate:
$\binom{6}{2} = 15$.
So:
$P(X = 2) = 15 \cdot \left(\frac{1}{2}\right)^{6} = 15 \cdot \frac{1}{64} \approx 0.2344$.
Step 3: Final Answer:
$P(X = 2) \approx 0.2344$.
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:
$P(King|Face) = \frac{1}{3}$.
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:
$P(X > 120) \approx 0.0918$.
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:
$E(X) = 4$.
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:
$Var(X) = 2.1$.
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:
$P(X < 10) = 0.5$.
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:
$Median = 5$.
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:
$P(\text{Rolling a 3 and Heads}) = \frac{1}{12}$.
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:
$Var(X) = 4$.
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:
$P(\text{Rolling a 1 or 2}) = \frac{1}{3}$.
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:
$P(25 < X < 35) \approx 0.6826$.
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:
$P(\text{Sample mean} > 53) \approx 0.0505$.
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).
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:
$E(X) = 4$.
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:
$Var(X) = 3$.
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:
$P(X < 5) \approx 0.6321$.
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 $B(\alpha, \beta) = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)}$.
Calculate:
$B(2, 5) = \frac{\Gamma(2) \Gamma(5)}{\Gamma(7)} = \frac{1 \cdot 24}{720} = \frac{1}{30}$.
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:
$f(0.5) \approx 0.46875$.
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 $3$.
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:
$F(1) = P(X \leq 1) = 0.3$.
$F(2) = P(X \leq 2) = 0.3 + 0.5 = 0.8$.
$F(3) = P(X \leq 3) = 0.3 + 0.5 + 0.2 = 1.0$.
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:
$P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$
$P(X = 2) = \frac{\binom{5}{2} \binom{3}{0}}{\binom{8}{2}}$.
Calculate:
$\binom{5}{2} = 10$,
$\binom{3}{0} = 1$,
$\binom{8}{2} = 28$.
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:
$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$
$P(X = 1) = \frac{2^1 e^{-2}}{1!} = 2 e^{-2}$.
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 $P(X \geq 5)$:
$P(X \geq 5) \approx P(Z \geq \frac{4.5 – 6}{2.19}) \approx P(Z \geq -0.68)$.
Use z-table:
$P(Z \geq -0.68) \approx 0.7517$.
Step 3: Final Answer:
$P(X \geq 5) \approx 0.7517$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 6) = \binom{8}{6} (0.5)^6 (0.5)^{2}$.
Calculate:
$\binom{8}{6} = 28$.
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:
$P(Ace|Not Ace) = \frac{P(Ace)}{P(Not Ace)} = \frac{4}{48} = \frac{1}{12}$.
Step 3: Final Answer:
$P(Ace|Not Ace) = \frac{1}{12}$.
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:
$P(X = 5) = \frac{32768 e^{-8}}{120}$.
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:
$Range_{<4} = 4 – 3 = 1$.
Calculate probability:
$P(X < 4) = \frac{Range_{<4}}{Range} = \frac{1}{4} = 0.25$.
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:
$P(X = 4) = \frac{1296 e^{-6}}{24}$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^{2} \left(\frac{5}{6}\right)^{2}$.
Calculate:
$\binom{4}{2} = 6$.
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:
$Range_{4-6} = 6 – 4 = 2$.
Calculate probability:
$P(4 < X < 6) = \frac{Range_{4-6}}{Range} = \frac{2}{6} = \frac{1}{3}$.
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:
$f(x) = \frac{x^{\alpha – 1} (1 – x)^{\beta – 1}}{B(\alpha, \beta)}$
$B(2, 5) = \frac{\Gamma(2)\Gamma(5)}{\Gamma(7)} = \frac{1 \cdot 24}{720} = \frac{1}{30}$.
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:
$f(0.3) \approx 1.441$.
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:
$Var(X) \approx 0.9868$.
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 $k = 10$:
$P(X = 9) = \binom{10}{9} (0.5)^9 (0.5)^{1} = 10 \cdot \frac{1}{1024} = \frac{10}{1024}$.
$P(X = 10) = \binom{10}{10} (0.5)^{10} = 1 \cdot \frac{1}{1024} = \frac{1}{1024}$.
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:
$P(X > 8) \approx 0.01074$.
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:
$P(X < 36) \approx 0.3085$.
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:
$Var(X) = 16$.
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:
$n = 10$,
$k \geq 7$.
Step 2: Solution:
Calculate probabilities for $k = 7, 8, 9, 10$.
$P(X = 7) = \binom{10}{7} (0.5)^{10} = \frac{120}{1024}$.
$P(X = 8) = \binom{10}{8} (0.5)^{10} = \frac{45}{1024}$.
$P(X = 9) = \binom{10}{9} (0.5)^{10} = \frac{10}{1024}$.
$P(X = 10) = \binom{10}{10} (0.5)^{10} = \frac{1}{1024}$.
Now add:
$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$
$= \frac{120 + 45 + 10 + 1}{1024} = \frac{176}{1024} \approx 0.1719$.
Step 3: Final Answer:
$P(X \geq 7) \approx 0.1719$.
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:
$P(\text{Heart}|\text{Red}) = \frac{1}{2}$.
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:
$P(\text{King and Queen}) = \frac{4}{663}$.
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:
$P(Red) = \frac{5}{12}$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 3) = \binom{5}{3} (0.4)^3 (0.6)^{2}$.
Calculate:
$\binom{5}{3} = 10$.
So:
$P(X = 3) = 10 \cdot (0.064) \cdot (0.36) = 10 \cdot 0.02304 \approx 0.2304$.
Step 3: Final Answer:
$P(X = 3) \approx 0.2304$.
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:
$E(X) = 2.2$.
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:
$F(1) = P(X \leq 1) = 0.2$.
$F(2) = P(X \leq 2) = 0.2 + 0.5 = 0.7$.
$F(3) = P(X \leq 3) = 0.2 + 0.5 + 0.3 = 1$.
Step 3: Final Answer:
CDF values are:
$F(1) = 0.2$,
$F(2) = 0.7$,
$F(3) = 1$.
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:
$P(X < 10) = 0.5$.
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:
$P(\text{> 4}) = \frac{1}{3}$.
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:
$Var(X) = 2.4$.
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:
$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$
$P(X = 2) = \frac{5^2 e^{-5}}{2!} = \frac{25 e^{-5}}{2}$.
Calculate:
$= \frac{25 \cdot 0.006737947}{2} \approx 0.0842$.
Step 3: Final Answer:
$P(X = 2) \approx 0.0842$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 1) = \binom{10}{1} (0.5)^1 (0.5)^{9}$.
Calculate:
$\binom{10}{1} = 10$.
So:
$P(X = 1) = 10 \cdot (0.5)^{10} = 10 \cdot \frac{1}{1024} = \frac{10}{1024} = \frac{5}{512}$.
Step 3: Final Answer:
$P(X = 1) = \frac{5}{512}$.
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:
$E(X) = 1.1$.
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:
$Range_{<10} = 10 – 5 = 5$.
Calculate probability:
$P(X < 10) = \frac{Range_{<10}}{Range} = \frac{5}{10} = 0.5$.
Step 3: Final Answer:
$P(X < 10) = 0.5$.
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 $6$.
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:
$P(X < 5) \approx 0.4647$.
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:
$P(\text{Heads and 3}) = \frac{1}{12}$.
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:
$Var(X) = 100$.
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:
$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$P(X = 4) = \binom{5}{4} (0.7)^4 (0.3)^{1}$.
Calculate:
$\binom{5}{4} = 5$.
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:
$P(X = 4) \approx 0.36015$.
