# probability of at least 1 girl

I am having trouble understanding how the probability of at least one girl is 3/4 in this question below.

Given a family of two children (assume boys and girls equally likely, that is, probability 1/2 for each), That at least one is a girl?

My reason for the confusion is that, if at least one is a girl then the probability of having a boy-boy does not exist, therefore the sample space becomes

[GG,BG,GB]

so surely the answer should be 2/3

• These are seperate questions. First what is the probability for two boys. Then what is the probability, that at least one child is a girl and so on. Aug 18, 2018 at 16:04
• "Both are girls" is the case $GG$. If you have $[GG,BG,GB],$ how do you conclude that the probability of $GG$ is $2/3$? Aug 18, 2018 at 16:06
• "... probability of at least one girl.. " in which situation? Please type the framework for only these problem, further related questions (with different context) make the question hard to find. Aug 18, 2018 at 16:07
• I will edit the question Aug 18, 2018 at 16:07
• The probability of "at least one girl" is $1-Prob(BB)=1-\frac 14=\frac 34$. You are confusing ordinary probability with conditional probability, though even then I don't see how you get $\frac 23$.
– lulu
Aug 18, 2018 at 16:10

Your sample space is not $[GG, BG, GB].$

Given a family of two children (assume boys and girls equally likely, that is, probability 1/2 for each) ...

This gives us (so far) the sample space $[GG, BG, GB, BB].$

... what is the probability ...

You left these words out of the question, but I think they were implied. They mean that we should now get ready to compute the probability of some event.

... that at least one is a girl?

OK, so the event is "at least one girl." The following elements of your sample space have at least one girl: $GG, BG, GB.$ The element $BB$ does not.

So the event we are asked to give the probability of is $[GG, BG, GB].$

The sample space is still $[GG, BG, GB, BB].$

We are evidently meant to assume that the births are independent as well as equally likely to be boys or girls, so the elements of the sample space are equally likely. Thus an event with $3$ of the $4$ elements has probability $3/4.$

The question does not assume there is at least one daughter in the family. The only assumption about children is there are two. Hence the prob. space is (BG, GB, BB, GG) and three of the four elementary events satisfy the requirement.

It isn't hard to list the cases. Since each has a probability of $\frac12,$ there are equally likely cases $GG,BG,GB,BB.$ Out of the four cases, $3$ have at least one girl, so it is $\frac34.$
Complementary counting also works. The probability of no girls (all boys) is $\left(\frac{1}{2}\right)^2=\frac14.$ The probability of at least one girl is $1-\frac14=\frac34.$
Either way, the probability is $\frac34.$
It seems you are answering the different problem

If there is at least one girl in a family with $2$ children, then what is the probability of only one girl?

• this is what I am not understanding, as how can you included in the probability the BB as there will never be a chance of that occurring. Aug 18, 2018 at 16:12
• It states at least one girl as the probability, not a given. Aug 18, 2018 at 16:16

The probability of having two boys is 1/4, not 0.

\begin{align} P(child1=boy)\times P(child2=boy) = \frac{1}{2}\times\frac{1}{2}\\ = \frac{1}{4}. \end{align}

No, equally likely either GB or BG so answer is 1/2