A couple plans to have three children. There are eight possible arrangements of

Question

A couple plans to have three children. There are eight possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All eight arrangements are (approximately) equally likely.

(a) What is the probability of any one of these arrangements?

(b) Let X be the number of girls the couple has. What is the probability that X= 2?

(c) Given that the first child born is a boy, what is the probability that the last 2 children are a boy and a girl?

Explanation / Answer

Solution:

A couple plans to have three children

( a )
all arrange ments are equally likely

probability of any one of these arrangements = 0.125

( b )

Let X be the number of girls the couple has.

combinations that has (X =2) are BGG,GBG,GGB

therefore, prpbability of (X = 2) is = 0.125 + 0.125 + 0.125

= 0.375

( c )

Given that the first child born is a boy

n = 2

p = 0.5

q = 0.5

binomial probaility distribution

Formula:

P(k out of n )= n!*pk * qn-k / k! *(n - k)!

P( X = 2 ) = 2!*0.52 * 0.52-2 / 2! *(2 - 2)!

= 0.125

BBB (1/2)*(1/2)*(1/2) 0.125 BGG (1/2)*(1/2)*(1/2) 0.125 BGB (1/2)*(1/2)*(1/2) 0.125 BBG (1/2)*(1/2)*(1/2) 0.125 GGG (1/2)*(1/2)*(1/2) 0.125 GBB (1/2)*(1/2)*(1/2) 0.125 GBG (1/2)*(1/2)*(1/2) 0.125 GGB (1/2)*(1/2)*(1/2) 0.125

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