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).

Step-by-step explanation:

The probability of 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 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 .

Alternative:

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

indicates the intersection of and . 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, . 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 .

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

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)