Question

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).

Answer 1

Answer:

$P(B|A)=\frac{2}{7}$

Step-by-step explanation:

The probability of $P(B|A)$ 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 $P(B|A)$ 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 $P(B|A)=\boxed{\frac{2}{7}}$.

Alternative:

For dependent events $A$ and $B$, the conditional probability of event B occurring given A is given by:

$P(B|A)=P(B\cap A)\div P(A)$

$P(B\cap A)$ indicates the intersection of $P(B)$ and $P(A)$. 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, $P(B\cap A)=\frac{2}{16}=\frac{1}{8}$. 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 $P(A)=\frac{7}{16}$.

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

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