Problema con calcolo della probabilità

Let's tackle this problem step by step. I'm going to break it down to understand the probability that a person Giovanni finds, who has exactly two children with at least one born on a Tuesday, has two boys.

Understanding the Problem
We have the following information:

A person has exactly two children.

At least one of the two children is born on a Tuesday.

The probability of a child being born on any given day of the week is equal (1/7 for each day).

The probability of a child being a boy or a girl is equal (1/2 for each).

We need to find the probability that both children are boys given these conditions.

Possible Combinations of Children
First, let's consider all possible combinations of genders and birth days for two children. Each child can be:

Boy (B) or Girl (G)

Born on any of the 7 days of the week (let's denote them as Mon, Tue, Wed, Thu, Fri, Sat, Sun)

So, for each child, there are 2 (genders) × 7 (days) = 14 possibilities.

For two children, the total number of possible combinations is 14 × 14 = 196.

However, since the order of the children matters (first child and second child are distinct), we don't need to adjust for order in this count.

Favorable Outcomes Where at Least One Child is a Boy Born on Tuesday
We're interested in cases where at least one of the two children is a boy born on Tuesday (let's denote this as BTu).

First, let's count all possible outcomes where at least one child is BTu.

This can be calculated as:
Total outcomes with at least one BTu = (Total outcomes where first child is BTu) + (Total outcomes where second child is BTu) - (Outcomes where both are BTu)

This is to avoid double-counting the case where both are BTu.

Number of outcomes where first child is BTu: 1 (BTu) × 14 (any for the second child) = 14

Number of outcomes where second child is BTu: 14 (any for the first child) × 1 (BTu) = 14

Number of outcomes where both are BTu: 1 (BTu) × 1 (BTu) = 1

So, total with at least one BTu = 14 + 14 - 1 = 27

Now, out of these 27 outcomes, how many have both children as boys?

Outcomes with Two Boys and At Least One BTu
We need both children to be boys, and at least one of them is BTu.

Possible cases:

First child is BTu, second child is a boy born on any day (Bany).

Second child can be BMon, BTu, BWed, BThu, BFri, BSat, BSun → 7 options

But if second is also BTu, that's included.

Second child is BTu, first child is a boy born on any day (Bany).

Similarly, first child can be any of the 7 boy-day combinations.

But again, we've double-counted the case where both are BTu.

So, total two-boy outcomes with at least one BTu:
= (First is BTu and second is any Bday) + (Second is BTu and first is any Bday) - (Both are BTu)
= 7 (second is any Bday) + 7 (first is any Bday) - 1 (both BTu)
= 7 + 7 - 1 = 13

Calculating the Probability
The probability that both children are boys given that at least one is a boy born on Tuesday is:
= (Number of favorable two-boy outcomes with at least one BTu) / (Total number of outcomes with at least one BTu)
= 13 / 27

Verifying with Enumeration
To ensure this is correct, let's enumerate the possibilities more carefully.

Each child can be represented as (Gender, Day). Let's list all combinations where at least one is BTu, i.e., at least one (B, Tue).

Total possible (Child1, Child2) where at least one is BTu:

First child is BTu, second child is anything:

BTu & (BMon, BTu, BWed, BThu, BFri, BSat, BSun, GMon, GTu, GWed, GThu, GFri, GSat, GSun) → 14

First child is not BTu but second child is BTu:

(BMon, BWed, BThu, BFri, BSat, BSun, GMon, GTu, GWed, GThu, GFri, GSat, GSun) & BTu → 13
(We exclude BTu in the first child here to avoid double-counting the case where both are BTu)

Total: 14 (first is BTu) + 13 (second is BTu, first not BTu) = 27

Now, how many of these have both children as boys?

From the first 14:

BTu & Bany: 7 (since Bany can be any of the 7 boy-day combinations)
From the next 13:

Bany & BTu: 6 (since first is B but not BTu, so BMon, BWed, BThu, BFri, BSat, BSun)
Total two-boy outcomes: 7 + 6 = 13

Thus, probability = 13 / 27.

Why Not 1/3?
You might initially think that since one child is a boy, the probability that the other is also a boy is 1/3 (from the classic two-child problem where possible siblings are BB, BG, GB). However, the additional information about the birth day adds more granularity, increasing the probability slightly because the "boy born on Tuesday" can be either child, and in the case of two boys, there's a higher chance that at least one is born on Tuesday compared to having one boy and one girl.

The probability that the person found by Giovanni has two boys, given that they have exactly two children and at least one is a boy born on a Tuesday, is 13/27.