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 boy

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).

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Adela 5 years 2021-07-19T03:47:06+00:00 1 Answers 47 views 0

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  1. Vodka
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    2021-07-19T03:48:20+00:00
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    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}}

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