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) ?0.530.040.140.16
Accepted Solution
A:
Answer: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) / 0.29P (S∩LC) = 0.552 × 0.29 = 0.16