ppennycuff405125
30.07.2021 •
Mathematics
13. A recent survey by the cancer society has shown that the probability that someone is a smoker is P(S) = 0.31. They have also determined that the probability that someone has lung
cancer, given that they are a smoker is P(LCS) = 0.226. What is the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer
P(SnLC)?
-0.08
-0.73
-0.25
-0.07
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Ответ:
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Work Shown:
S = person is a smokerLC = person has lung cancerP(S) = 0.31 = probability someone is a smokerP(LC given S) = probability someone has lung cancer, given they are a smokerP(LC given S) = 0.226Use that given info to say the following:
P(LC given S) = P(LC and S)/P(S)
P(LC and S) = P(LC given S)*P(S)
P(LC and S) = 0.31*0.226
P(LC and S) = 0.07006
P(LC and S) = 0.07
This problem is an example of using conditional probability.
I used "and" in place of the intersection symbol
Saying P(LC and S) is the same as P(S and LC). The order doesn't matter.
Ответ:
gfuo
Step-by-step explanation: