Question

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

Answer 1

Answer: Choice D) 0.07

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Work Shown:

• S = person is a smoker
• LC = person has lung cancer
• P(S) = 0.31 = probability someone is a smoker
• P(LC given S) = probability someone has lung cancer, given they are a smoker
• P(LC given S) = 0.226

Use 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
`\cap`

Saying P(LC and S) is the same as P(S and LC). The order doesn't matter.