Question

A survey of athletes at a high school is conducted, and the following facts are discovered: 33% of the athletes are football players, 66% are basketball players, and 20% 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?

Answer 1

Answer:   0.79 or 79%

Step-by-step explanation:

Given : The probability that the athletes are football players : P(A)=0.33

The probability that the athletes are basketball players: P(B)=0.66

The probability that the athletes play both football and basketball : P(A∩B)=0.20

Now, the probability that the athlete is either a football player or a basketball player is given by :-

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\=0.33+0.66-0.20=0.79$

Hence, the probability that the athlete is either a football player or a basketball player = 0.79

Or we can write it as 79%.

Answer 2

Answer:

99%.

Step-by-step explanation:

Givens

• 33% of the athletes are football players.
• 66% are basketball players.
• 20% of the athletes play both football and basketball.

Now, these percentages are representations regarding the total number of students.

It's important to know that percentages can also represents probabilities, like in this case.

So, basically, there's 33% to chose a football player, there's 66% to choce a basetball player, and there's 20% to choce a student who plays both football and basketball.

Now, if we want to know the probability of choosing someone who plays either football or baskebatll, then we must sum 33% + 66% = 99%.

Therefore, the probability is 99% of picking a student who plays either football or basketball.