7. Boy and girl. Let's assume that the birth of a girl and a boy is equally likely. It is known that in a certain family there are two children.

a) What is the probability that one is a boy and one is a girl

b) It is additionally known that one of the children is a boy. What is the probability now that there is one boy and one girl in the family

c) It is additionally known that the boy was born on a Monday. What is the probability now that there is one boy and one girl in the family

![](https://cdn.mathpix.com/cropped/2024_05_06_ade23fa47a7a5f9f8672g-09.jpg?height=614&width=494&top_left_y=1075&top_left_x=1475)
Solution. a) Children appear in a certain sequence (MM, MD, DM, or DD). All sequences are equally likely, and the probability of each is $\frac{1}{4}$. The condition "Boy and girl" is favorable to two outcomes, MD and DM, so the probability of this is $\frac{2}{4}=\frac{1}{2}$.

b) Now it is known that one of the children is a boy. Out of the four sequences, three remain equally likely: MM, MD, and DM. Therefore, the probability of the event "Boy and girl" is now $\frac{2}{3}$.

c) Now it is known that the boy was born on a Monday. Let's introduce the days of the week into consideration: $M$ - Monday and O - other day (any day except Monday). Now we have sixteen possible sequences, nine of which

MMOO, MDOO, MDMO, DMOO, DDOO, DDMM, DMMO, DMDM, DMMO are impossible, as they do not contain the fragment MM (a boy born on a Monday). However, seven sequences containing MM are possible:

Considering that the probability of being born on a Monday is $\frac{1}{7}$, and on another day is $\frac{6}{7}$, we find that the total probability of the possible sequences is

$$
\begin{gathered}
\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{6}{7}+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{6}{7}+\frac{1}{2} \cdot \frac{6}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+ \\
+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+\frac{1}{2} \cdot \frac{6}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}=3 \cdot \frac{1}{4} \cdot \frac{1}{49}+4 \cdot \frac{1}{4} \cdot \frac{6}{49}=\frac{27}{196}
\end{gathered}
$$

Of these sequences, only four

$$
\text { MMDO, MMDI, DMMO, and DMMI }
$$

favor the event "boy and girl." Their total probability is

$$
\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{6}{7}+\frac{1}{2} \cdot \frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}+\frac{1}{2} \cdot \frac{6}{7} \cdot \frac{1}{2} \cdot \frac{1}{7}=2 \cdot \frac{1}{4} \cdot \frac{1}{49}+2 \cdot \frac{1}{4} \cdot \frac{6}{49}=\frac{14}{196}
$$

Therefore, the conditional probability of the event "boy and girl" given the condition "there is a boy born on a Monday" is

$$
\frac{14}{196} : \frac{27}{196} = \frac{14}{27}
$$

Answer: a) $\frac{1}{2} ;$ b) $\frac{2}{3} ;$ c) $\frac{14}{27}$.

Comment. The problem can be solved in different ways. For example, one can consider a simplified probability space where the elementary outcomes are events of the form MM*, M*MD, MM* and M*MM, where the asterisk denotes any day of the week. There are 7 events of each type, but one (MM*) is counted twice. Therefore, there are 27 events in total, and all are equally likely. Out of these, 14 events involve a girl, which leads to the result. It is somewhat surprising that the day of the week on which the boy was born changes the probability of a mixed-gender pair. This idea is somewhat unexpected, I agree.