According to a recent survey, outside of their own family members, 26% of adult Americans have no close friend to confide in. If this is the prevailing probability today, find the probability that in a random sample of n = 5 adults

(a) two or more have no close friend.
(b) at most two have no close friend.
(c) Find the expected number of persons who have no close friend.

Solution

Here $X$ denote the number of adult Americans who have no close friend to confide.

$p$ be the probability that adult American who have no close friend to confide.

Given that $p=0.26$ and $n =5$. Thus$X\sim B(5, 0.26)$.

The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{5}{x} (0.26)^x (1-0.26)^{5-x},\\ &\quad x=0,1,\cdots, 5. \end{aligned} $$

(a) The probability that 2 or more have no close friend is

$$ \begin{aligned} P(X\geq 2) & =1-P(X\leq 1)\\ &= 1-\sum_{x=0}^{1} P(x)\\ &=1-\big(P(X=0)+P(X=1)\big)\\ &= 1- (0.2219+0.3898)\\ & = 1-0.6117 \\ & = 0.3883 \\ \end{aligned} $$

(b) The probability that at most two have no close friend is

$$ \begin{aligned} P(X\leq 2) & =\sum_{x=0}^{2} P(x)\\ & =\sum_{x=0}^{2}\binom{5}{x}(0.26)^x(1-0.26)^{5-x}\\ &= P(X=0)+P(X=1) + P(X=1)\\ & = 0.2219+0.3898+0.2739\\ &=0.8857. \end{aligned} $$

(c) The expected number of persons who have no close friend is $E(X) = n*p = 5 * 0.26 = 1.3$.

Further Reading