# Find the probability that a student has siblings of a certain gender

A coworker of mine has a seventh grader who got a probability question wrong on a test, a couple of us at work are looking at it and we're getting the same answers that he got.

This is the question verbatim:

A seventh grader has 2 siblings. 1) What is the probability that this seventh grader is a girl with a brother and a sister?

2) What is the probability that there are at least 2 girls in the family?

For the first question we're getting either 1/8 or 1/6. We get 1/8 by 1/2*1/2*1/2 = p(student is girl) * p(bro) * p(sis). We get 1/6 by 1/2*1/3 = p(student is girl) * p(last 2 siblings are a boy and a girl)

For the second we're getting 1/2. We list out the combinations and half of them meet this criteria (BBB, GGG, BGG, GBB).

None of our answers are correct according to the teacher. Can someone help us out here?

• I agree with Simon's answer for part 1. You could also get this by writing out all 8 possible families and see that 2 of the 8 meet the criteria. But it seems like 1/2 is the answer for part two (although not the combinations you enumerated as examples. it would be BGG, GGB, GBG, GGG) – turkeyhundt Mar 26 '15 at 17:59
• Additionally, using your logic that got you 1/8 for part 1, to that you would have to add p(student is girl) * p(sis) * p(bro) to get the 1/4 – turkeyhundt Mar 26 '15 at 18:01

The probability that the two sibling are both boys is $(1/2)^2 = 1/4$. Likewise that both are girls. The probability that one is a boy and one is a girl is $2\times (1/2)^2 = 1/2$.

Hence the probability the seventh grader is a girl (probability = $1/2$) and has 2 siblings who are a boy and girl is $1/2 \times 1/2 = 1/4$.

If there are three children, and each is either a boy or a girl, if there are two or more girls there are one or fewer boys, but if there are fewer than two girls there are at least two boys. So "2 boys" and "2 girls" are mutually exclusive and cover all possibilities. If each child is equally likely to be a boy or a girl, by symmetry the two events have equal probability, $1/2$.

Under the same assumptions, the probability of exactly two of three children is a girl is $3/8$ (three different ways for there to be exactly one boy in a sequence of three, each with $1/8$ probability).

Given that the chosen seventh-grader has two siblings, you have chosen a child from a three-child family, and the probability that there are exactly two girls in the family is $3/8$. Given those facts, the probability that the child you chose was one of the girls in that family is $2/3$. So the chance that you chose a girl from a three-child family with exactly two girls is $\frac38 \cdot \frac23 = 1/4.$

1) What is the probability that this seventh grader is a girl with a brother and a sister?

For the first question we're getting either 1/8 or 1/6. We get 1/8 by 1/2*1/2*1/2 = p(student is girl) * p(bro) * p(sis).

That would work if you wanted the probability that the eldest child was a girl, the middle child a boy, and the youngest child a girl. But your criteria doesn't specify an order.

There are two ways the other two siblings could be a boy and a girl. They could be a boy then a girl, or a girl then a boy. So the probability must be multiplied by 2.

\begin{align} \mathsf P(S=g\cap F=\{g,g,b\}) & = \mathsf P(S=g)\;\mathsf P(F\setminus\{S\}=\{g,b\}) \\[1ex] & = \frac 1 2\cdot\frac 2 4 \\[1ex] & = \frac 1 4 \end{align}

Now, can you answer the second question?