Two students are selected at random from a group consisting of five girls andthree boys. Find the probability that both students are girls, given that atleast one is a girl.

To solve this problem, we need to use conditional probability. Let's denote the events as follows:

• $A$: Both students selected are girls.
• $B$: At least one student selected is a girl.

We are looking for $P(A|B)$, the probability that both students are girls given that at least one is a girl. According to the definition of conditional probability:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

First, let's find $P(A)$, the probability that both students selected are girls.

The total number of ways to select 2 students out of 8 (5 girls + 3 boys) is given by the combination formula $\binom{n}{k}$:

$\binom{8}{2} = \frac{8!}{2!(8-2)!} = 28$

The number of ways to select 2 girls out of 5 is:

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10$

So, the probability that both students are girls is:

$P(A) = \frac{\binom{5}{2}}{\binom{8}{2}} = \frac{10}{28} = \frac{5}{14}$

Next, let's find $P(B)$, the probability that at least one student selected is a girl. We can use the complement rule here. The complement of event $B$ (at least one girl) is the event that no girls are selected (both students are boys).

The number of ways to select 2 boys out of 3 is:

$\binom{3}{2} = 3$

So, the probability that no girls are selected is:

$P(B^c) = \frac{\binom{3}{2}}{\binom{8}{2}} = \frac{3}{28}$

Therefore, the probability that at least one student is a girl is:

$P(B) = 1 - P(B^c) = 1 - \frac{3}{28} = \frac{25}{28}$

Now, we need to find $P(A \cap B)$. Since event $A$ (both students are girls) is a subset of event $B$ (at least one girl), $P(A \cap B) = P(A)$:

$P(A \cap B) = P(A) = \frac{5}{14}$

Finally, we can find $P(A|B)$:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{14}}{\frac{25}{28}} = \frac{5}{14} \times \frac{28}{25} = \frac{10}{25} = \frac{2}{5}$

So, the probability that both students are girls, given that at least one is a girl, is $\frac{2}{5}$.