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ivolga24 [154]
3 years ago
15

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 deter

mined 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
Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

<u>Answer:</u>

The correct answer option is P (S∩LC) = 0.16.

<u>Step-by-step explanation:</u>

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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