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A couple has three children. Two events are stated. One is the occurrence of a third boy. The other is the occurrence of the first two girls.

The probability of occurrence of an event B given that an event A has already occurred is referred to as theconditional probability of B given A.

It has the following formula:

$P\left(B|A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

The number of possible outcomes on any birth is two, girls or boys, such that each is equally likely.

Therefore, the probability of a girl or boy on any given birth would be 0.5 $\left(\frac{1}{2}\right)$.

As each child is independent to be born, the probability that the third child is a boy would remain the same as 0.5.

The mathematical proof for the same can also be shown as follows:

Let A be the event of having the first two girls.

Let B be the event of having a third boy.

The sample space for three children is given as follows:

S = {bbb,bbg,bgb,gbb,bgg,gbg,ggb,ggg}

Here, b represents a boy, whereas g represents a girl.

The total number of outcomes in the sample space is 8.

The outcome 鈥済gb鈥� depicts the occurrence of A and B together.

Thus, the probability of having a third boy and first two girls is given by:

$P\left(AandB\right)=\frac{1}{8}$

The outcomes 鈥済gb and ggg鈥� depict the occurrence of A.

Thus, the probability of having the first two girls is given by:

$P\left(A\right)=\frac{2}{8}$

The probability of B, given A, is computed as follows:

$$\begin{gathered} P\left(B|A\right)=\frac{P\left(AandB\right)}{P\left(A\right)} \\ =\frac{\frac{1}{8}}{\frac{2}{8}} \\ =\frac{1}{2} \\ =0.5 \end{gathered}$$

Therefore, the probability of having a third boy, given that the first two are girls, is equal to 0.5.