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rewona [7]
3 years ago
15

According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women

are randomly selected: a. What is the probability that exactly 2 of them have never been married? b. That at most 2 of them have never been married? c. That at least 13 of them have been married?
Mathematics
1 answer:
blagie [28]3 years ago
8 0

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : P(X=r) =^nC_r p^r q^{n-r}

At x = 2

P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}

P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}

P(X=2) =8.8009 \times 10^{-7}

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So,  P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}

 P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})

P(X=r) =9.9439 \times 10^{-6}

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=0.2360

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