# Thread: Probability of a girl?

1. ## Probability of a girl?

Q: Find the probability of a couple having a baby girl when their fourth child is born, given that the first 3 children are girls.

A: I think the answer is 0.5. Because regardless of the situation the probability will always be 0.5 as each event is independent.

How can I solve this problem in terms of P(B|A)=P(A and B)/P(A)

2. Hello, crazydeo!

Good question!

Q: Find the probability of a couple's fourth child is a girl,
given that their first 3 children are girls.

A: I think the answer is 0.5, because regardless of the situation,
the probability will always be 0.5, as each event is independent.

How can I solve this problem in terms of: . $P(B|A)\:=\:\frac{P(A \wedge B)}{P(A)}$

We want: . $P(\text{4th is girl }|\text{ first 3 are girls}) \;=\;\frac{P(\text{[4th is girl] }\wedge\text{ [first 3 are girls]})}{P\text{(first 3 are girls})}$

The numerator is: . $P(\text{all 4 are girls}) \;=\;\left(\frac{1}{2}\right)^4 \:=\:\frac{1}{16}$

The denominator is: . $P(\text{first 3 are girls}) \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8}$

Therefore: . $P(\text{4th is a girl }|\text{ first 3 are girls}) \;=\;\frac{\frac{1}{16}}{\frac{1}{8}} \;=\;\frac{1}{2}$

3. Just a comment on the problem--

Both crazydeo and Soroban assume that the genders of the children are (statistically) independent. I think this is very likely true, but it is not automatically so. One can imagine biological mechanisms which would result in non-independence.

(According to this web page

Houghton Mifflin Science for Families: Cricket Connections

there was an article in "Chance" which concluded there is no compelling evidence for non-independence.)

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# the possibility of a couple having a daughter as a fourth child is

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