So the question looks like this A survey finds that 30% of all Canadians enjoy jogging. Of those Canadians who enjoy jogging, it is known that 40% of them enjoy swimming. For the 70% of Canadians who do not enjoy jogging, it is known that 60% of them enjoy swimming. Suppose a Canadian is randomly selected. What is the probability that they enjoy exactly one of the two activities? Select the answer closest to yours. I solved this using $$ P(Jogging or Swimming)=P(Jogging) + P (Swimming) - P(Jogging and swimming) P(Jogging)=0.3 P(Swimming)=0.54 P(Jogging and swimming)=0.12 $$ my answer was 0.72 but it was wrong. Please, what am I missing?

kadirsmr9d 2022-09-14 Answered
So the question looks like this A survey finds that 30% of all Canadians enjoy jogging. Of those Canadians who enjoy jogging, it is known that 40% of them enjoy swimming. For the 70% of Canadians who do not enjoy jogging, it is known that 60% of them enjoy swimming.
Suppose a Canadian is randomly selected. What is the probability that they enjoy exactly one of the two activities? Select the answer closest to yours.
I solved this using P(JoggingorSwimming)=P(Jogging)+P(Swimming)P(Joggingandswimming)P(Jogging)=0.3P(Swimming)=0.54P(Joggingandswimming)=0.12
my answer was 0.72 but it was wrong. Please, what am I missing?
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Answers (1)

acilschoincg8
Answered 2022-09-15 Author has 12 answers
Your answer calculates the probability that the selected person enjoys either swimming or jogging, and possibly both. The question is asking for the probability that the person enjoys exactly one of the activities, not both.
The probability of this is P(swimming or jogging) - P(swimming and jogging). So you should get .72−.12=.6
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