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galben [10]
3 years ago
7

A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla

yers, 52% are basketball players, and 9% of the athletes play both football and basketball. An athlete is chosen at random from the high school: what is the probability that he is either a football player or a basketball player?
Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = P(A)= \frac{13}{100}=0.13

The probability that athlete is a basketball player = P(B)= \frac{52}{100}=0.52

The probability that athlete is both basket ball player and  football player = P(A\cap B) = \frac{9}{100}=0.09

We have to find the probability that an athlete chosen is either a football player or a basketball player P(A\cup B).

Now we know that;

P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

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