6.23 In the town of Hickoryville, an adult citizen has a 1% chance of being selected for jury duty next year. If jury members are selected at random, and the Anderson family includes three adults, determine the probability that

a. none of the Andersons will be chosen.
b. exactly one of the Andersons will be chosen.
c. exactly two of the Andersons will be chosen.

Solution

Let $X$ denote the number of members selected for jury out of $n=3$. An adult citizen has a 1% chance of being selected for jury duty next year, i.e. $p = 0.01$.

The probability distribution of $X$ is Binomial distribution. That is $X\sim B(3,0.01)$.

The probability mass function of $X$ is

$$ \begin{aligned} P(X=x) &= \binom{3}{x} (0.01)^x (1-0.01)^{3-x},\\ &\qquad \; x=0,1,\cdots, 3 \end{aligned} $$

a. The probability that none of the Andersons will be chosen is

$$ \begin{aligned} P(X= 0) & =\binom{3}{0} (0.01)^{0} (1-0.01)^{3-0}\\ & = 0.9703. \end{aligned} $$

b. The probability that exactly one of the Andersons will be chosen is

$$ \begin{aligned} P(X= 1) & =\binom{3}{1} (0.01)^{1} (1-0.01)^{3-1}\\ & = 0.0294. \end{aligned} $$

c. The probability that exactly two of the Andersons will be chosen is

$$ \begin{aligned} P(X= 2) & =\binom{3}{2} (0.01)^{2} (1-0.01)^{3-2}\\ & = 0.0003. \end{aligned} $$