Question

A store ran an advertisement for a sale in a newspaper and on TV. The probability a resident saw the ad in the newspaper is 0.13. The probability a resident saw the ad on TV is 0.34. Assume whether a customer saw the ad in the newspaper is independent of whether the person saw the ad on TV. What is the probability a resident saw the ad in the newspaper given the person saw the ad on TV?

Answer 1

Answer: Our required probability is 0.13.

Explanation:

Since we have given that

Probability that a resident saw the ad in the newspaper P(N) = 0.13

Probability that a resident saw the ad on TV P(T) = 0.34

Since they are independent events.

So,
`P(N\cap T_=P(N).P(T)`

So, Probability that a resident saw the ad in the newspaper given the person saw the ad on TV is given by

`P(N|T)=(P(N\cap T))/(P(T))\\\\P(N|T)=(P(N).P(T))/(P(T))\\\\P(N|T)=P(N)=0.13`

Hence, our required probability is 0.13.