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The family has two children. We know that one is Jack(boy). What is the probability that the other child is also boy? The ratio of the number of newborn boys and girl 107:100 (boys are more).

My solution:

If is one boy, we have 4 options:

Jack-girl 1/4=0,25

girl-boy 1/4=0,25

Jack-boy 1/4=0,25

boy-Jack 1/4=0,25

Only two options can be, in this case is probability 2/4=0,5

The probability of the birth of a boy - 107/(107+100) = 0,5169

The probability of the birth of a girl - 100/(107+100) = 0,4831

The probability that two boys are born for each other is 0,5169*0,5169

The finally probability that the second child is also a boy 0,5*0,5169*0,5169 = 0,1336

Is this solution correct?


marked as duplicate by Namaste discrete-mathematics Jan 12 '17 at 16:46

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  • $\begingroup$ I think you found the case when the family has 3 children - read the question again. $\endgroup$ – mdave16 Jan 12 '17 at 16:41
  • $\begingroup$ I think it is just $\frac{107}{207}$. Imagine this: you are running into the mother of this family in a store, and she has Jack with her, but not the other child. When talking to her, she tells you that Jack is her son, and that she has a second child as well, but she doesn't say what gender it is. OK, so in that scenario, you know that there is this family with two children, one of which is Jack, which is compatible with your givens. So, what is the chance the other child is a boy (assuming randomness, etc.) ? It would simply be the chance of any one child being a boy, i.e. $\frac{107}{207}$. $\endgroup$ – Bram28 Jan 12 '17 at 17:19
  • $\begingroup$ Yes, It is different from the standard 'if a family has two children, and at least one is a boy, what are the chances the other is a boy?' But I think that barak manos solution is correct $\endgroup$ – Lama Jan 12 '17 at 18:02
  • $\begingroup$ I suppose it depends on how you know that this family has two children with one a boy called Jack. If you run into the mother with Jack by her side in the store and she tells you she has one more kid, it's $\frac{107}{207}$. But if you are told: 'OK, Mrs. Wilson has two children, one of which is a boy called Jack', then it's $\frac{107}{307}$ (following barak manos solution below) $\endgroup$ – Bram28 Jan 12 '17 at 20:19
  • $\begingroup$ @Lama Given my earlier comments, my answer is: we don't have enough information to answer this question, as it depends on what 'we know that Jack (a boy) is one of them' really means. In particular, we don't know in which way we know that Jack is one of the children. $\endgroup$ – Bram28 Jan 12 '17 at 21:59

No, you have been given a condition that one of them is a boy. You need to apply Bayes' theorem here, which relates the conditional probabilities.


Unfortunately it is not correct; Have you ever heard of the gambler's fallacy? This situation is pretty similar to what happens in a casino:

If you go to a casino and you pass over the roulette, and you notice that before you got there, the roulette rolled $15$ black numbers in a row, wouldn't you be tempted to bet everything on red? Many people would, but that is wrong! Why??

The roulette doesn't remember what numbers already got out; Every time you roll it, it does not care about what has already been rolled. The odds never change; There are still 18 black numbers and 18 red numbers, thus the probability of rolling a red number will always be $\frac12$ regardless of how many black numbers already came out in a row.

The same thing for pregnant women. You can ask yourself this: if I am to get 2 kids, what are the odds of having 2 boys? If the boy-girl ratio is $50:50$, then that would be $\frac12\cdot\frac12 = \frac14$. Assume one boy was born. What is the probability of having a second boy? Well, when the mother gets pregnant, there is a $50\%$ chance she is carrying a boy, $50\%$ chance she is carrying a girl, so there is $50\%$ chance you are going to have 2 boys! That probability would be expressed as "What are the odds of having 2 boys, given that I have 1 boy already": this is what we call a conditional probability, which is $P(A | B)$, read as "the probability of happening $A$ given that $B$ happened", and even has some nice formula:

$$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

For that problem statement, you must calculate

$$P(\text{getting a second boy} |\ \text{I have one boy already})$$

Which, by the formula, becomes:

$$P(\text{getting a second boy} |\ \text{I have one boy already}) = \\ \frac{P(\text{having a second boy}\cap\text{had one boy first})}{P(\text{had one boy first})}$$


Let A denote the probability of $2$ boys.

Let B denote the probability of at least $1$ boy.

Then $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\frac{107}{207}\cdot\frac{107}{207}}{\frac{100}{207}\cdot\frac{107}{207}+\frac{107}{207}\cdot\frac{107}{207}+\frac{107}{207}\cdot\frac{100}{207}}=\dfrac{107}{307}\approx34.85\%$.


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