It has been reported that 70% of college graduates did some sort of volunteer work last year. If 4 college graduates were randomly selected, and x is the number who did volunteer work, determine the following:

1. Probability that none of the selected graduates did volunteer work.
2. Probability that exactly one of the selected graduates did volunteer work.
3. Probability that two or more of the selected graduates did volunteer work.
4. Probability that between one and three of the selected graduates did volunteer work.

Answer:
Given data:

The probability of success, $p = 0.70$
Sample size, $n = 4$

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(a) The indicated probability can be calculated by considering binomial distribution as:

$P(x = 0) = (nCx) * p^x * (1 – p)^{n – x} $

$P(x = 0) = (4C0) * 0.7^0 * (1 – 0.7)^{4 – 0}$

$P(x = 0) \approx 0.0081$

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(b) The indicated probability can be calculated by considering binomial distribution as:

$P(x = 1) = (nCx) * p^x * (1 – p)^{n – x} $

$P(x = 1) = (4C1) * 0.7^1 * (1 – 0.7)^{4 – 1}$

$P(x = 1)= 0.0756$

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(c) The indicated probability can be calculated by considering binomial distribution as:

$P(x \geq 2) = 1 – P(x \leq 1)$

$P(x \geq 2) = 1 – \Sigma_{x=0}^{1} (4Cx) * 0.7^x * (1 – 0.7)^{4 – x}$

$P(x \geq 2) = 1 – 0.0837$

$P(x \geq 2) = 0.9163$

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(d) The indicated probability can be calculated by considering binomial distribution as:

$P(1 \leq x \leq 3) = \Sigma_{x=1}^{3} (nCx) * p^x * (1 – p)^{n – x}$

$P(1 \leq x \leq 3) = \Sigma_{x=1}^{3} (4Cx) * 0.7^x * (1 – 0.7)^{4 – x}$

$P(1 \leq x \leq 3) = 0.7599 – 0.0081$

$P(1 \leq x \leq 3) = 0.7518$