Math

QuestionWhat is the probability that a randomly chosen athlete plays football or basketball, given 45%45\% are football players, 25%25\% are basketball players, and 19%19\% play both? Enter as a whole number percentage.

Studdy Solution

STEP 1

Assumptions1. The percentage of athletes who are football players is45% . The percentage of athletes who are basketball players is25%
3. The percentage of athletes who play both football and basketball is19%
4. We are looking for the probability that an athlete chosen at random is either a football player or a basketball player

STEP 2

To find the probability that an athlete is either a football player or a basketball player, we need to add the probabilities of each event. However, since some athletes play both sports, they are counted twice. Therefore, we need to subtract the probability of an athlete playing both sports.
(FB)=(F)+(B)(FB)(F \cup B) =(F) +(B) -(F \cap B)Where- (FB)(F \cup B) is the probability of an athlete being a football player or a basketball player- (F)(F) is the probability of an athlete being a football player- (B)(B) is the probability of an athlete being a basketball player- (FB)(F \cap B) is the probability of an athlete being both a football player and a basketball player

STEP 3

Now, plug in the given values for the probabilities.
(FB)=45%+25%19%(F \cup B) =45\% +25\% -19\%

STEP 4

Perform the calculation.
(FB)=45%+25%19%=51%(F \cup B) =45\% +25\% -19\% =51\%So, the probability that an athlete chosen at random is either a football player or a basketball player is51%.

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