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

Question

Mathematics

Guest

24 January, 11:05

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

+3

Answers (1)

Know the Answer?

Ignacio Morgan · Answer 1

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