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Flauer [41]
3 years ago
6

Twelve percent of U.S. residents are in their forties. Consider a group of nine U.S. residents selected at random. Find the prob

ability that two or three of the people in the group are in their forties.
Mathematics
1 answer:
miskamm [114]3 years ago
7 0

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\%

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