TITLE: The family has two children. We know that one is Jack(boy). What is the probability that the other child is also boy? Newborn ratio 107:100
QUESTION [0 upvotes]: 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?

REPLY [1 votes]: 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:
$$P(\text{getting a second boy} |\ \text{I have one boy already}) = \\
\frac{P(\text{having a second boy}\cap\text{had one boy first})}{P(\text{had one boy first})}$$

REPLY [0 votes]: 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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