Question

Problem: Suppose that a couple will have 3 children. Assume that the probability of having a girl is .487 and the probability of having a boy is .513. Also assume that x is a random variable for the number of girls and find the probability where x is 0, 1, 2, or 3. In order words, answer the following probability questions.

What is the probability that none of the three children will be girls?

What is the probability that exactly one of the three will be girls?

What is the probability that exactly two of the three will be girls?

What is the probability that all three of the children will be girls?

Answer 1

a)

Here, n = 3, p = 0.487, (1 - p) = 0.513 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 3C0 * 0.487^0 * 0.513^3
P(X = 0) = 0.135

b)

Here, n = 3, p = 0.487, (1 - p) = 0.513 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 1)
P(X = 1) = 3C1 * 0.487^1 * 0.513^2
P(X = 1) = 0.3845
0

c)

Here, n = 3, p = 0.487, (1 - p) = 0.513 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 2)
P(X = 2) = 3C2 * 0.487^2 * 0.513^1
P(X = 2) = 0.365

d)

Here, n = 3, p = 0.487, (1 - p) = 0.513 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 3C3 * 0.487^3 * 0.513^0
P(X = 3) = 0.1155
0