Question

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?

Answer 1

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)

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

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

$=0.2360$

Answer 2

A: The probability for 2 women who have never been married in the random selection of 15 women is $8.80 \times 10^{-7}$.

B: The probability for at most 2 women who have never been married in the random selection of 15 women is $9.9 \times 10^{-6}$.

C: The probability for at least 13 women who have never been married in the random selection of 15 women is 0.176.

What is Probability?

Probability is a numerical description of how likely an event is to occur, or how likely it is that a proposition is true. The higher the probability of an event, the more likely it is that the event will occur.

Given that 75% of employed women have never been married. We need to find the probability for women never married if 15 women are randomly selected.

This can be calculated with the formula for binomial distribution is probability.

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

Where P is the binomial probability, x is the number of times for a specific outcome within n trials, p is the probability of success on a single trial, q is the probability of failure on a single trial, n is the number of trials and ${^nC_x}$ is the number of combinations.

For the given situation, p = 75 %= 0.75, q = 25%= 0.25, n = 15

Part A: The probability that exactly 2 of them have never been married.

x = 2, Then the probability is,

$P(2) = ^{15}C_2\times (0.75)^2 \times (0.25)^ {15-2}$

$P(2) = \dfrac { 15!}{2! (15-2)!}\times 0.5625 \times (0.25)^{13}$

$P(2) = 8.80\times 10^{-7}$

The probability for 2 women who have never been married in the random selection of 15 women is $8.80 \times 10^{-7}$.

Part B: At most 2 of them have never been married.

x = 0, 1, 2, Then the probability is,

$P(0 to 2) = ^{15}C_0\times (0.75)^0 \times (0.25)^ {15-0} + ^{15}C_1\times (0.75)^1 \times (0.25)^ {15-1} + ^{15}C_2\times (0.75)^2 \times (0.25)^ {15-2}$

$P(0to 2) = 1\times 1\times 1 + 15 \times 0.75 \times 3.72 \times 10^{-9} + 8.80 \times 10^{-7}$

$P(0to 2) = 9.9 \times 10^{-6}$

The probability for at most 2 women who have never been married in the random selection of 15 women is $9.9 \times 10^{-6}$.

Part C: At least 13 of them have been married.

x = 13, 14, 15, Then the probability is,

$P(13 to 15) = ^{15}C_{13}\times (0.75)^{13} \times (0.25)^ {15-13} + ^{15}C_{14}\times (0.75)^{14} \times (0.25)^ {15-14} + ^{15}C_{15}\times (0.75)^{15} \times (0.25)^ {15-15}$

$P(13 to 15) = \dfrac {15!}{13! \times (15-13)!}\times 0.75^{13} \times 0.25^2 + \dfrac {15!}{14! \times (15-14)!}\times 0.75^{14} \times 0.25^1 + \dfrac {15!}{15! \times (15-15)!}\times 0.75^{15} \times 0.25^0$

$P(13to15) = 0.1559 + 0.0068 + 0.0133$

$P(13to15) = 0.176$

The probability for at least 13 women who have never been married in the random selection of 15 women is 0.176.

To know more about the probability, follow the link given below.

https://brainly.com/question/795909.