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Johnny has 4 children. It is known that he has more daughters than sons. Find the probability that the last child is a boy.

I let A be the event that the last child is a boy, P(A) = $\frac{1}{2}$. and B be the event that he as more daughters than sons. But im not sure how to calculate P(B) and what are the subsequent steps to take after.

Appreciate any help.

Thanks

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    $\begingroup$ +1 for asking a simple question and getting three different answers (already deleted) :) $\endgroup$ – Peter Franek Nov 25 '14 at 7:44
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    $\begingroup$ I guess we are to assume that his decisions on whether to have another child are independent of the genders of his existing children, but this assumption should be stated. It could be that he (and presumably his partner) kept having children until they had at least one boy and at least one girl, and then stopped. In that case, given that they have more girls than boys, the last one is definitely a boy. $\endgroup$ – Robert Israel Nov 25 '14 at 8:30
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    $\begingroup$ I'd say its 50% it's gonna be a boy $\endgroup$ – DevEstacion Nov 25 '14 at 9:23
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    $\begingroup$ We do need to know about it, because it affects the probabilities, just as the statement that he has more girls than boys does. $\endgroup$ – Robert Israel Nov 25 '14 at 16:06
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    $\begingroup$ Wouldn't it be great to have statistical information from several countries and being able to verify the real numbers? And checking if they are the same if we exchange "boy" and "girl"? $\endgroup$ – gnasher729 Nov 25 '14 at 18:11
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If he has more daughters than sons, Below are the 5 possible cases:

D D D D --> All Daughters
S D D D --> 3 Daughters
D S D D
D D S D
D D D S

So probability of having last child as son is = 1/5.

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    $\begingroup$ One should also add that all of these options are equally likely, but that holds if we have no other information then in the assignment. $\endgroup$ – Peter Franek Nov 25 '14 at 7:43
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    $\begingroup$ @PeterFranek Excellent remark, from which I would rather be inclined to draw a different conclusion than yours: if we have no other information in the assignment, the assignment is faulty. $\endgroup$ – Did Nov 25 '14 at 7:56
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    $\begingroup$ +1. People tend to forget that probability is fundamentally about counting. $\endgroup$ – Pete Becker Nov 25 '14 at 14:54
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    $\begingroup$ Pardon me for passing thru, non mathematician here, but isn't the probability always just 50%, it's either a boy or a girl? But if the odds are 1/5 it's a boy, does that mean it's 4/5 (or 80%) chance it will be a girl? $\endgroup$ – Madivad Nov 25 '14 at 21:18
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    $\begingroup$ @Madivad The probability of a child being a boy or a girl is equally 50% only if we have no context, but as you can see this problem provides a lot of context and correctly understanding the context will lead you to the solution. $\endgroup$ – Alin Purcaru Nov 25 '14 at 23:52
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The number of girls in the family would have a binomial distribution, so the prior probability that there are 3 or 4 girls in the family would be:

$$\begin{align} \mathsf P(B) & = {4\choose 3}(\tfrac 1 2)^3(\tfrac 1 2)+{4\choose 4}(\tfrac 1 2)^4 \\ & = \frac 5{16} \end{align}$$

Now for the probability that the last child in the family is a boy and that there are more girls than boys in the family is equal to: the prior probability that the first three children are girls and the last is a boy:

$$\begin{align} \mathsf P(A\cap B) & = \frac{1}{16} \end{align}$$

Thus the posterior probability, that the last child is a boy given that their are more girls in the family than boys is:

$$\begin{align} \mathsf P(A\mid B) & = \frac{\mathsf P(A\cap B)}{\mathsf P(B)} \\ & = {\frac 1 {16}}\bigg/\frac 5 {16} \\ & = \dfrac 1 5 \end{align}$$

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There are $2^4=16$ possible permutations of children, e.g. MMMM, or MFFM, or FFFM, or FFFF (here order is important, hopefully it's clear that MFFF means the first child is male, the second is female, and so on) and each is equally likely. Now it just becomes conditional probability.

Let $A$ be the event that the last child is male. Let $B$ be the event that there are more female children than male children. The probability we are looking for is $P(A|B)$, the probability of $A$ given $B$. This is given by $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ where $A\cap B$ is the event where both $A$ and $B$ occur. Let's look at the terms individually:

  • $A\cap B$ occurs precisely when Johnny has more daughters than sons, and the last child is a son. It's not hard to see there's only one possibility here: FFFM. Hence $P(A\cap B)=\frac1{16}$.
  • $B$ occurs either when Johnny has no sons (one option, FFFF) or one son (four options, MFFF, FMFF, FFMF and FFFM). There are five options in total, so $P(B)=\frac5{16}$.

Putting this all together we get $$P(A|B)=\frac{\frac1{16}}{\frac5{16}}=\frac15$$ so the probability that Johnny's last child was a son is one in five.

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The probability that any birth is a boy or a girl is NOT 1:1 as many people believe; in actuality 105 boys are born for every 100 girls. This ratio of 1.05 is known as the "secondary sex ratio." Given these real world statistics, one must give 1.05 weight to the four scenarios that include one boy and 1.00 weight to the all-girl possibility.

Therefore, the answer would be 1.05/5.20, or 21/104.

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    $\begingroup$ This is completely irrelevant for the problem, which starts “Johnny has four children”. $\endgroup$ – egreg Nov 26 '14 at 8:06
  • $\begingroup$ I modified my post to include an answer, in addition to showing why the real world probability is indeed relevant. $\endgroup$ – wildBillMunson Nov 26 '14 at 8:21
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    $\begingroup$ This answer isn't incorrect per se, but the solver clearly isn't expected to know the precise gender ratio in probability exercises of this nature. $\endgroup$ – COTO Nov 26 '14 at 8:36
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Picking up Robert Israel's suggestion: If a couple keeps having children until they have at least a boy and a girl, and ignoring twins, ... it's difficult.

If they had children until they reached their goal, the probability is 1/16 each that they stop at 3 boys and 1 girl, at 1 boy and 3 girls, 3/4 that they stopped with 2 or 3 children, 1/8 that they had five or more.

But they may not be finished yet. They may have four sons or four daughters and be waiting for another one. After being married for some time, there is a probability p that they have enough time for four children, and a probability q < p that they have enough time for five children. Both p and q grow with time, but don't reach 1.

The probabilities if there are four children: p/16 for 1B + 3G, p/16 for 3B + 1G, (p - q)/16 each for 4B or 4G. Since there are more girls, p/16 for 1B + 3G and the last is a boy, (p-q)/16 for four girls. With conditional probabilities, the probability that the last is a boy is p / (2p - q).

It would depend on the length of the relationship. If they are together for four years, it would be quite unlikely that they could have five children, (q would be small compared to p) and the probability of GGGB would be only slightly larger than GGGG, so the probability would be only a bit higher than 0.5. If they are together for many years, having GGGG and no fifth child is unlikely, so the chance that the last one is a boy is closer to 1.

That also means that statistics for currently living parents and historical data would show different numbers.

Obviously for different parental strategies the result would be different.

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    $\begingroup$ If their strategy was to keep having children until they have one of each (or figure they have enough already), then there are only two possible sequences (ignoring twins triplets etc). GGGG and GGGB. Each is equally likely, within a small margin of error. Thus probability that the last child is a boy, given the information in the post plus this assumption, is about 0.5. $\endgroup$ – Theodore Norvell Nov 26 '14 at 1:13
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They are mutually exclusive events. How many son/daughters the couple already has is irrelevant. Hence, the probability of the last child (or rather any child) being boy is 1/2 (some would say 1/3 too).

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    $\begingroup$ I assume you mean "independent" rather than "mutually exclusive"? No. The gender of the next child may reasonably be assumed to be independent of those that came before, but the gender of the latest child is not independent of the more girls than boys statement. $\endgroup$ – Rawling Nov 26 '14 at 8:11

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