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:

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

if i put something wrong your not gonna like it so rather not say nothing.