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22 April, 09:22

A recent survey by the cancer society has shown that the probability that someone is a smoker is P (S) = 0.29. They have also determined that the probability that someone has lung cancer, given that they are a smoker is P (LC|S) = 0.552. What is the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer P (S∩LC) ?

0.53

0.04

0.14

0.16

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  1. 22 April, 10:49
    0
    The correct answer option is P (S∩LC) = 0.16.

    Step-by-step explanation:

    It is known that the probability if someone is a smoker is P (S) = 0.29 and the probability that someone has lung cancer, given that they are also smoker is P (LC|S) = 0.552.

    So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P (S∩LC).

    P (LC|S) = P (S∩LC) / P (S)

    Substituting the given values to get:

    0.552 = P (S∩LC) / 0.29

    P (S∩LC) = 0.552 * 0.29 = 0.16
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