# Solved:69% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself

#### ByDr. Raju Chaudhari

Sep 26, 2020

69% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan. Find the probability that the number who consider themselves baseball fans is

(a) exactly five,
(b) greater than or equal to six, and
(c) less than four.

#### Solution

Here $X$ denote the number of men who consider themselves professional baseball fans.

$p$ be the probability that men who consider themselves professional baseball fans.

Given that $p=0.69$ and $n =10$. The random variable $X\sim B(10, 0.69)$.

The probability mass function of $X$ is
$$P(X=x) = \binom{10}{x} (0.69)^x (1-0.69)^{10-x}, \; x=0,1,\cdots, 10.$$

(a) The probability that the number who consider themselves baseball fans is exactly five is

 \begin{aligned} P(X= 5) & =\sum_{x=0}^{5} P(x)-\sum_{x=0}^{4} P(x)\\\\ & = 0.1128\\ \end{aligned}

(b) The probability that the number who consider themselves baseball fans is greater than or equal to six is
 \begin{aligned} P(X\geq 6) & =\sum_{x=6}^{10} P(x)\\ & =\sum_{x=6}^{10}\binom{10}{x}(0.69)^x(1-0.69)^{10-x}\\ & = \binom{10}{6} (0.69)^{6} (1-0.69)^{10-6}+\binom{10}{7} (0.69)^{7} (1-0.69)^{10-7}\\ & \quad +\binom{10}{8} (0.69)^{8} (1-0.69)^{10-8}+\binom{10}{9} (0.69)^{9} (1-0.69)^{10-9}\\ &\quad +\binom{10}{10} (0.69)^{10} (1-0.69)^{10-10}\\ & = 0.2093 +0.2662+0.2222+0.1099+0.0245\\ & = 0.8321 \end{aligned}

(c) The probability that the number who consider themselves baseball fans is less than four is

 \begin{aligned} P(X< 4) & =\sum_{x=0}^{3} P(x)\\ & =\sum_{x=0}^{3}\binom{10}{x}(0.69)^x(1-0.69)^{10-x}\\ & = \binom{10}{0} (0.69)^{0} (1-0.69)^{10-0}+\binom{10}{1} (0.69)^{1} (1-0.69)^{10-1}\\ & \quad +\binom{10}{2} (0.69)^{2} (1-0.69)^{10-2}+\binom{10}{3} (0.69)^{3} (1-0.69)^{10-3}\\ & = 0 +0.0002+0.0018+0.0108\\ & = 0.0129 \end{aligned}