Question

According to government data, 46% 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

Solution

Given that ,

p = 0.46

1 - p = 0.54

n = 15

a)

x = 2

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X = 2) = ((15! / 2! (15 - 2)!) * 0.46² * (0.54)^15-2

=  ((15! / 2! (13)!) * 0.46² * (0.54)^15-2

= 0.0774

Probability = 0.0774

b)

x 2

P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)

= ((15! / 0! (15 - 0)!) * 0.46⁰ * (0.54)^15-0 + ((15! / 1! (15 - 1)!) * 0.46¹ * (0.54)^15-1 + ((15! / 2! (15 - 2)!) * 0.46² * (0.54)^15-2

= ((15! / 0! (15)!) * 0.46⁰ * (0.54)¹⁵ + ((15! / 1! (14)!) * 0.46¹ * (0.54)¹⁴ + ((15! / 2! (13)!) * 0.46² * (0.54)¹³

= 0.0016 + 0.0172 + 0.0774

Probability = 0.0963

c)

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

= ((15! / 13! (15 - 13)!) * 0.46¹³ * (0.54)^15-13 + ((15! / 14! (15 - 14)!) * 0.46¹⁴ * (0.54)^15-14 + ((15! / 15! (15 - 15)!) * 0.46¹⁵ * (0.54)^15-15   

= 0.0013 + 0.0002 + 0.0000

= 0.0015

Probability = 0.0015