# Solved (Free): According to the American lung association, 90% of adult smokers started smoking before turning 21 years old. Ten smokers 21 years old or older

#### ByDr. Raju Chaudhari

Mar 28, 2021

According to the American lung association, 90% of adult smokers started smoking before turning 21 years old. Ten smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.
Find the probability that exactly 3 of them started smoking before 21 years of age.
Find the probability that at least 3 of them started smoking before 21 years of age.

#### Solution

Here $X$ denote the number of smokers who started smoking before 21 is recorded out of selected 10.

Let $p$ be the probability that an adult smokers started smoking before turning 21 years old.

Given that $p=0.9$ and $n =10$. Thus $X\sim B(10, 0.9)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{10}{x} (0.9)^x (1-0.9)^{10-x},\\ &\quad x=0,1,\cdots, 10. \end{aligned}

$x$ $P(X=x)$
0 0.00000
1 0.00000
2 0.00000
3 0.00001
4 0.00014
5 0.00149
6 0.01116
7 0.05740
8 0.19371
9 0.38742
10 0.34868

The probability that exactly $3$ of the calls involve a fax message is

 \begin{aligned} P(X= 3) & =\binom{10}{3} (0.9)^{3} (1-0.9)^{10-3}\\\\ & = 0.00001\\ \end{aligned}

The probability that at least $3$ of them started smoking before 21 years of age is

 \begin{aligned} P(X\geq 3) & =1-P(X\leq 2)\\ &= 1-\sum_{x=0}^{2} P(x)\\ &= 1- 0\\ & = 1 \end{aligned}