On a multiple-choice mid-term exam for an online Statistics course, one of the questions read:
A couple plans to have three children. The chance they have a girl is 0.8. The gender of one child is independent of the gender of another child. What is the probability of having one boy and two girls when having three children?
The choices were:
a) 0.008
b) 0.096
c) 0.384
d) 0.512
It's the only question I answered incorrectly but I won't be able to discuss it with my instructor for a while so I decided to ask here for someone to explain why my answer is incorrect and show me the correct answer. Here is my approach:
If the chance they have a girl is 0.8, then the chance they don't have a girl (they have a boy) is 1 - 0.8 = 0.2. Therefore, the probability of having one boy and two girls out of three children is 0.2 * 0.8 * 0.8 = 0.124. However, 0.124 wasn't one of the choices so I chose b) 0.096, the choice closest to 0.124.
Firstly, .2 *.8 * .8 = .128, not .124
.128 is the probability that they will have: first a boy, then 2 girls.
You need to realize they are asking for a combination, not a permutation.
You can have a boy and 2 girls in three different orders: BGG, GBG, GGB. The probability you are looking for is 3*0.128= 0.384
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