Question

According to a study, 90 % of adult smokers started smoking before 21 years old. 11 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

1. The probability that at least 7 of them started smoking before 21 years of age is
2. The probability that at most 3 of them started smoking before 21 years of age is
3. The probability that exactly 3 of them started smoking before 21 years of age is

Answer 1

Solution:

Given that, n = 11, p = 0.90

Now let by using binomial distribution P(X=x) = nCx * p^x q^(n-x) we solve below probabilities.

1.

We have to find the probability P(X>=7) = P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11)  
P(X>=7) = 11C7 * (0.9)^7 * (0.1)^4 + 11C8 * (0.9)^8 * (0.1)^3 + 10C9 * (0.9)^9 * (0.1)^2 + 11C10 * (0.9)^10 * (0.1)^1 + 11C11 * (0.9)^11 * (0.1)^0 = 0.6834

2.

P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0 + 0 + 0 + 0 = 0

3.

P(X = 3) = 11C3 * (0.9)^3 * (0.1)^9 = 0

Best Luck !