What is the expected number of boys?

Question

If the probability that a child is a boy is $p$, find the expected number of boys in a family with $n$ children given that there is at least one boy?

My answer:

My friend says the answer is $np+q$ but my answer is $np/(1-q^n)$. I found my answer through conditional probability by finding the $P(X=k/X>0)$ .

My friend tells that we can find the answer by finding the expected number of boys for $n-1$ children and then add $1$ to the expectation. Which one is right?

Computing the expected number of boys for $n-1$ children and then adding 1 will not give the right answer, unless you are conditioning on the fact that the first child is a boy. — user432944, Sep 27 '18 at 14:44
q=1-p, and we are asked to find the mean that is, expectation — user90596, Sep 27 '18 at 16:04
I am not closing this question because it is still not logically clear why my friends method is wrong... — user90596, Sep 27 '18 at 16:43
I understand it intuitively due to the answer provided by @EspeciallyLine, but still I don't understand what cinch exists in that logic and where exactly it went wrong — user90596, Sep 27 '18 at 16:53

score 4 · Answer 1 · answered Sep 27 '18 at 16:08

4

Let $X$ represent the number of boys $($successes$)$ out of $n$ children $($trails$)$.

So, $X$ follows the binomial distribution.

The expected number of boys in a family with $n$ children given that there is at least one boy is $$E(X/X\ge 1)=\sum_{x}x\cdot P(X=x/X\ge1)$$$$=0+\sum^n_{x=1}x\cdot\dfrac{P(X=x)}{P(X\ge1)}$$ $$=\sum^n_{x=1}\dfrac{x\cdot P(X=x)}{1-P(X<1)}$$ $$=\sum^n_{x=1}\dfrac{x\cdot P(X=x)}{1-P(X=0)}$$ $$=\sum^n_{x=1}\dfrac{x\cdot P(X=x)}{1-^nC_{0}P^0q^{n-0}}$$ $$=\dfrac{1}{1-q^n}\sum^n_{x=1}x\cdot P(X=x)$$ $$=\dfrac{1}{1-q^n}E(X)$$ Therefore, $$E(X/X\ge1)=\dfrac{1}{1-q^n}np=\dfrac{np}{1-q^n}$$

answered Sep 27 '18 at 16:08

Key Flex

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Thanks, finally thanks.,.... – user90596 Sep 27 '18 at 16:09
Could you tell what's wrong with the other answer, that is adding 1 to expectation for n-1 children – user90596 Sep 27 '18 at 16:10
@user90596 Without the work, I couldn't tell where the mistake is – Key Flex Sep 27 '18 at 16:11
But that's it the working is: – user90596 Sep 27 '18 at 16:12
E(X) for n-1 children + 1 – user90596 Sep 27 '18 at 16:12
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@user90596 $np+q=(n-1)p+1$ would be the expectation if you knew that the youngest child (say) was a boy. Merely knowing that at least one child is a boy is less information, so it should change the expectation by a smaller amount. – Especially Lime Sep 27 '18 at 16:13
Could you please explain in a bit more detail? – user90596 Sep 27 '18 at 16:17
Do you mean that the 1 child we added might have been a girl? – user90596 Sep 27 '18 at 16:19

callculus42 · Answer 2 · 2018-09-27T16:43:11.297

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You can use the law of total expectation.

$$\mathbb E(X)=\mathbb E(X|X=0)\cdot P(X=0)+\mathbb E(X|X>0)\cdot P(X>0)$$

It should be obvious that $E(X|X=0)=0$. Thus

$$\mathbb E(X|X>0)=\frac{\mathbb E(X)}{P(X>0)}$$

Now we use the converse probability: $P(X>0)=1-P(X=0)=1-(1-p)^n$. Consequently

$$\mathbb E(X|X>0)=\frac{n\cdot p}{1-(1-p)^n}$$

edited Sep 27 '18 at 16:43

answered Sep 27 '18 at 16:27

callculus42

30,550

Could you tell how my friend's method is wrong? – user90596 Sep 27 '18 at 16:44
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@user90596 I don´t see the logic behind the method of your friend. At the expectation $p(n-1)$ there is the case $X=0$ included But this is not $E(X|X<n)$. If I´m right $$E(X|X<n)=\frac{np(1-p^{n-1})}{1-p^n}$$ – callculus42 Sep 27 '18 at 17:12
Ok, I will think about this.. – user90596 Sep 27 '18 at 17:17
Ok. It is a good idea to sleep on the problem for a night. – callculus42 Sep 27 '18 at 17:20
Still, I can't understand. Oh no...... – user90596 Sep 27 '18 at 17:46
@user90596 The problem is that your friend hasn´t given a sufficient explanation why he/she first calculate the expectation for $n-1$ (and then add $1$). But as I said, in the expectation of $n-1$ the case $X=0$ is included. But this case should be anyhow excluded. – callculus42 Sep 27 '18 at 17:55
So, if we calculate using that method for n=2 the expectation turns out to be 1.5 while through our method the answer turns out to be 1.33 which value do you think is more practical? – user90596 Sep 27 '18 at 18:01
I understand that X=0 should be excluded – user90596 Sep 27 '18 at 18:03
But when we add 1 it IS excluded – user90596 Sep 27 '18 at 18:03
"So, if we calculate using that method for n=2 the expectation turns out to be 1.5 while through our method the answer turns out to be 1.33 which value do you think is more practical?" $1.\overline 3=1\frac13$ is right. "But when we add 1 it IS excluded" Why it is something excluded if you add something? – callculus42 Sep 27 '18 at 18:16
We don't exclude it, we consider that it surely exists and does not need to have it's probability calculate d – user90596 Sep 27 '18 at 18:26
@user90596 I have to sleep on the solution of yor friend for a night as well. I´m not very confident that it helps to understand the aproach, but you never know. – callculus42 Sep 27 '18 at 18:47
Thanks for putting so much effort – user90596 Sep 27 '18 at 18:53

antkam · Accepted Answer · 2018-09-27T21:50:40.920

Others have already explained why your answer is correct. This answer hopefully illustrates why your friend is wrong. Your friend probably has the following common confusion on what conditional probability means -- I had the same confusion 30+ years ago. :)

Let $n=2, p=1/2$. There are 4 cases $\{BB, BG, GB, GG\}$ where the first letter denotes the sex of the older child, and the second letter denotes the sex of the younger child. Each case has equal probability $1/4$.

Problem stmt: "given there is at least one Boy" means condition on the event $E= \{BB, BG, GB\}$. The expected number of Boys is ${1 \over 3} (2 + 1 + 1) = {4 \over 3}$.
Now consider a different problem stmt: "given the older child is a Boy". This would condition on the event $E'=\{BB, BG\}$ and the expected no. of Boys $={1\over 2} (2 + 1) = {3 \over 2}$. This is your friend's answer.

Basically your friend is counting a "sure thing" (the $1$) and then assuming that the sure thing does not affect the distribution of "the rest" (the $(n-1)p$). If the "sure thing" is the older boy, then indeed that does not affect the distribution of the rest of the kids. But if the "sure thing" is "at least one boy" then it does affect the distribution of the rest of the kids, because who constitutes "the rest" may change depending on who constitutes the "sure thing".

A simpler version of this example: Flip 2 fair coins -- and for dramatic effect use 2 distinguishable coins (e.g. a quarter + a dime, if American). If there is at least 1 Head, what is the probability the other coin is also Head? The correct answer is $1/3$, while your friend would wrongly answer $1/2$. If your friend insists, I suggest you offer 50-50 odds and bet on the other coin being Tail (correct prob $2/3$). Remember to physically illustrate the meaning of conditional probability as follows: whenever the result is 2 Tails (i.e. not at least 1 Head), then it's a do-over and you reflip both coins. Play this 20 times, win some money, and convince your friend.

What is the expected number of boys?

3 Answers3