There are 16 kids on a soccer team. The players are either boys or girls, and they either prefer playing offense or defense. There are 6 boys who prefer offense, 3 boys who prefer defense, 5 girls who prefer offense, and 2 girls who prefer defense.

Suppose you choose one player from the team at random. Let event A be that the player is a girl, and event B be that the player prefers defense. Fine P(B | A).

The probability of [tex]P(B|A)[/tex] can be read as the probability of event B occurring given event A. In this question, event A occurs when the chosen player is a girl. There are 7 girls on the soccer team. Event B occurs when the chose player plays defense. Since [tex]P(B|A)[/tex] stipulates that event A already occurred, we want the probability of choosing a player who prefers defense from the 7 girls. There are 2 girls who prefer defense, hence [tex]P(B|A)=\boxed{\frac{2}{7}}[/tex].

Alternative:

For dependent events [tex]A[/tex] and [tex]B[/tex], the conditional probability of event B occurring given A is given by:

[tex]P(B|A)=P(B\cap A)\div P(A)[/tex]

[tex]P(B\cap A)[/tex] indicates the intersection of [tex]P(B)[/tex] and [tex]P(A)[/tex]. In this case, it is the probability that both events occur. Since there are 16 kids on the soccer team and only 2 are girls and prefer defense, [tex]P(B\cap A)=\frac{2}{16}=\frac{1}{8}[/tex]. The probability of event A occurring (chosen player is a girl) is equal to the number of girls (7) divided by the number of kids on the team (16), hence [tex]P(A)=\frac{7}{16}[/tex].

Therefore, the probability of event B occurring, given event A occurred, is equal to:

[tex]P(B|A)=\frac{1}{8}\div \frac{7}{16},\\\\P(B|A)=\frac{1}{8}\cdot \frac{16}{7}=\boxed{\frac{2}{7}}[/tex]