Q:

Twelve percent of U.S. residents are in their forties. Consider a group of nine U.S. residents selected at random. Find the probability that two or three of the people in the group are in their forties.

Accepted Solution

A:
Answer:The probability is 28 % ( approx )Step-by-step explanation:Since, there are only two possible outcomes in every trails ( residents are in their forties or not ),So, this is a binomial distribution,Binomial distribution formula,Probability of success in x trials,[tex]P(x)=^nC_x p^x q^{n-x}[/tex]Where, [tex]^nC_x=\frac{n!}{x!(n-x)!}[/tex]p and q are probability of success and failure respectively and n is the total number of trials.Let X be the event of resident who are in his or her forties.Here, p = 12% = 0.12,β‡’ q = 1 - p = 1 - 0.12 = 0.88,n = 9,Hence, the probability that two or three of the people in the group are in their forties = P(X=2) + P(X=3)[tex]^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}[/tex][tex]=36(0.12)^2 (0.88)^7+84(0.12)^3 (0.88)^6[/tex][tex]=0.279266611163[/tex][tex]\approx 0.28[/tex][tex]=28\%[/tex]