According to a survery 52% of the residents of a city oppose a downtown casino. Of these 52% about 5 out of 10 strongly oppose the casino.

AThe probability that a randomly selected resident opposes the casino and strongly opposes the casino is?

BThe probability that a randomly selected resident who opposes the casino is?

Question

13-02-2015
Mathematics

Answered

According to a survery 52% of the residents of a city oppose a downtown casino. Of these 52% about 5 out of 10 strongly oppose the casino.

AThe probability that a randomly selected resident opposes the casino and strongly opposes the casino is?

BThe probability that a randomly selected resident who opposes the casino is?

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Vuk15 Vuk15 · Answer 1 · 2017-03-14T10:20:41-04:00

I need to satert by stating the questions clearly. Question A: What is the probability that a radomly selected resident opposes the casino and strongly opposes the casino? This is the joint probability of being opposed and being strongly opposes which is calculated as the product of both: 52/100 5/10 = 26/100 = 0.26. Question B. What tis the probability that a radomly selected resident opposes the casino and does not strongly oppose the casino. Again it is a joint probability, the product of being opposed and not being strongly opposed: 52/100 5/10 = 26/100 = 0.26. The are the same because the the half of the 52% oppose strongly and the other half does not.