Question

A survey of athletes at a high school is conducted, and the following facts are discovered: 40% of the athletes are football players, 59% are basketball players, and 13% of the athletes play both football and basketball. An athlete is chosen at random from the high school: what is the probability that the athlete is either a football player or a basketball player? Probability = % (Please enter your answer as a percent)

Answer 1

Answer:

86%

Step-by-step explanation:

Let

A= Football player

B= Basket ball player

Football players=40%

Basketball players=59%

Both football and basket ball players=13%

Total percent=100%

The probability that athlete is a football player=P(A)=$\frac{40}{100}=0.40$

The probability that athlete is a basketball player=P(B)=$\frac{59}{100}=0.59$

The probability that athlete is both basket ball player and  football player=$P(A\cap B)=\frac{13}{100}=0.13$

We have to find the probability that athlete is either a football player or a basketball player.

It means we have to find $P(A\cup B)$

We know that

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A\cup B)=0.40+0.59-0.13=0.86=0.86\times 100=$86%

Hence, the probability that the athlete is either a football player or a basketball player=86%