Question

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.

Answer 1

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,

$P(x)=^nC_x p^x q^{n-x}$

Where, $^nC_x=\frac{n!}{x!(n-x)!}$

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,

⇒ 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)

$^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}$

$=36(0.12)^2 (0.88)^7+84(0.12)^3 (0.88)^6$

$=0.279266611163$

$\approx 0.28$

$=28\%$