Tricky Children Probability Question

Question

Guys I've been struggling with this problem for quite sometime and still cannot solve. Please help me!

Here's the question: We know that a couple has two children, and one of them is a boy born in a weekend. What is the probability that the other child is a girl?

Thanks in advance!

Here's what I've done so far:

Considering the days and genres, I think it looks like this: 2 cases: 1- 1 child boy born in weenked and second boy or girl born son Mon to Sun = 28 possibilities 2-2 ° son boy who was born in weekend and first boy born or is Mon to Friday or girl is born Mon to Sun = 24 possibilities

So we have 52 possibilities, and of these 28 are the case where one child is born in weekend and 1 girl born, so the prob = 28/52 = 7/13

PS: Sorry for just posting the question without my approach.

Let 1 represent boy and 0 represent girl. If a family has two children, then they are either 00, 10, 01, or 11. Assuming that the probability of each case is the same, you know that the boy belongs to the (11,10,01) group. Now what is the probability that the other number is 0? — polfosol, Oct 24 '16 at 05:21
@polfosol there is more information that you aren't using, namely the born on a weekend part. The question is a bit more complicated than what you make it out to be. See below. — JMoravitz, Oct 24 '16 at 05:42
Possible duplicate of Boy Born on a Tuesday - is it just a language trick? — Ian Miller, Oct 24 '16 at 05:50
This question (or ones very similar) have been asked many times before. Please use the search feature of StackExchange. — Ian Miller, Oct 24 '16 at 05:51

score 1 · Answer 1 · answered Oct 24 '16 at 05:32

You have to specify carefully how you pick the families. What you are expected to assume is that each child is a boy or girl with probability $\frac 12$ for each and that children are born on the weekend with probability $\frac 27$ regardless of sex. Your sample space is then all two child families that have at least one child who is a boy born on the weekend.

Why do I make a deal of this? As stated, the couple could have more than two children. You could have chosen to only look at couples with one boy and one girl. There might be a correlation between birth on the weekend and sex.

Given the assumptions in the first paragraph, there are three kinds of children: boys born on the weekend, who are $\frac 17$ of all children and we will call $A$, boys not born on the weekend, who are $\frac 5{14}$ of all children and we will call $B$, and girls, who are $\frac 12$ of all children an we will call $G$. We are told that we have two children and one is an $A$, so the choices in order are $AA, AB, AG, BA, GA$ with a priori probabilities $\frac 1{49}, \frac 5{98}, \frac 1{14}, \frac 5{98}, \frac 1{14}$. Scale these up to sum to $1$ and find what fraction are $AG$ or $GA$

score 1 · Accepted Answer · answered Oct 24 '16 at 05:41

First, we need to be perfectly clear about what is being asked.

Assuming we are looking only at families with two children, the gender of each child is uniformly distributed and independently at random, and the weekday of birth is uniformly distributed and independently at random.

Let $X$ represent the random variable which counts the number of male children of the couple. That is, $Pr(X=0)$ corresponds to the probability that both children are female, $Pr(X=1)$ corresponds to the probability that one is female and the other is male (in no specific order), and $Pr(X=2)$ is the probability that both children are male.

We have $Pr(X=0)=Pr(X=2)=\frac{1}{4}$ and $Pr(X=1)=\frac{1}{2}$

Let $W$ represent the event that at least one of the children is a boy born on a weekend.

We are asked to calculate $Pr(X=1\mid W)$

By Baye's theorem

$$Pr(X=1\mid W) = \frac{Pr(W\mid X=1)\cdot Pr(X=1)}{Pr(W)}$$

Given that we know there is exactly one boy, we see that to have a boy born on a weekend it must be that that specific boy was born on a weekend, which occurs with probability $\frac{2}{7}$ (assuming you count only saturday/sunday as weekend)

We already noted $Pr(X=1)=\frac{1}{2}$ from before.

Finally, we need to calculate $Pr(W)$. Note that $Pr(W)=Pr(W\cap ((X=0)\cup (X=1)\cup (X=2))) = Pr(W\cap (X=0)) + Pr(W\cap (X=1)) + Pr(W\cap (X=2))$

$=0 + Pr(X=1)Pr(W\mid X=1) + Pr(X=2)Pr(W\mid X=2)$

$=0+\frac{1}{2}\cdot \frac{2}{7} + \frac{1}{4}\cdot (1-(\frac{5}{7})^2)$

$=\frac{1}{7} + \frac{6}{49} = \frac{13}{49}$

So, we have:

$$Pr(X=1\mid W) = \frac{\frac{2}{7}\cdot \frac{1}{2}}{\frac{13}{49}} = \frac{7}{13}$$

(note: this agrees with the answer from Ross Millikan once you finish the necessary computations)

Tricky Children Probability Question

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