Paradox of the two children

The Paradoxe of the two children is one of most known problems of the Theory of probability.

Statement

“Knowing that a family has two children and that one of them is a boy, which is the probability that the other is a boy too? ”

This statement misses rigor at the point being ambiguous and one can give him two different interpretations which do not have the same answer: 1/3 for one and 1/2 for the other:

Is a group of parents of two children. I ask all those which have at least a son to form a sub-group. If I request from one parents of the sub-group if his/her other child is as a boy, which is the probability as he answers yes? Answer: 1/3.
: Explanation: In the group of parents, we have 1/4 parents of girl-girl (FF), 1/4 of girl-boy (FG), 1/4 of boy-girl (GF) and 1/4 of boy-boy (GG). Thus 3/4 of parents (FG, GF and GG) go in the sub-group. Only 1/3 of the parents of the sub-group has also a boy for another child, that is to say two boys (GG). Let us note that 3/4 × 1/3 = 1/4 and one finds the share of GG in the total.
I arrive to a knowledge of which I know only that it has two children. A child opens the door to me, it is a boy. Which is the probability that the second child is a boy? Answer: 1/2.
: Explanation: Before opening the door, I know that the family contains FF, FG, GF or GG. After having opened the door, I know that it contains GF or GG and probability that the second child is G is 1/2. If it is a girl who had opened, then I would know that the family contains FF or FG.
: On the other hand, if there was a habit requiring that a son open the door, in so far as there is a son, then the probability that the second child is also a son would be of 1/3. Indeed, in this case, after a boy opened the door to me, then I would know that I am in case FG, GF or GG because a girl can open only in the case FF. In case FG, the girl must yield her place to the boy, which influences the probabilities.

Defects of the statement

One cannot answer because the procedure by which information is obtained is not specified.

In a general way, if has and B are two event S (definite without ambiguity) of a probabilized space (he also nonambiguous), there is no ambiguity to determine the conditional probability of has knowing B.

On the other hand, if has is an event and B a Assertion, one cannot answer a exercise of the form “One knows B; which is the probability of has? ” In the simplest cases, one can try to replace this question within the framework “conditional probability of has knowing B” by interpreting the assertion B like an event B.

It as should be noted as random pulling is that of the family comprising two children, and that it is revealed then that one of the children is a boy (for better rendering comprehensible than one is not in the situation " a boy opens the porte" , one should rather say " the children are not two filles"). It is not the same thing as to say as one randomly draws a boy resulting from a family with two children.

Implicit assumptions

To solve this problem, we make the assumption that the probability of having a boy is of 50%. In reality, it is not completely the case. Us make as assumption as the births are independent events, which is not completely true, would be this only in the case of the Jumeaux monozygotes.

Formalization with two events

Is With the event: the two children are boys
Is B the event: one of both is a boy

Formally, one with the following equality: P (HAS|B) P (B) = P (B|A) P (A)

The probability of knowing B multiplied by that of B is equal to the probability of B knowing multiplied by that of A.

Thus P (has|B) = P (B|A) × P (A)/P (B) (see Theorem of Bayes)

However P (B|A) = 1 because if both are boys, then one of both is a boy (always true)

P (A) = 1/4 (idem demonstration above)

P (B) = 3/4 (idem demonstration above)

Thus P (has|B) = 1 × 4/1/3 /4 = 1/3

The probability that both are boys, knowing that one of both in is one, is of 1/3.

See too

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