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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?

Respuesta :

Answer: Our required probability is 0.13.

Step-by-step 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, [tex]P(N\cap T_=P(N).P(T)[/tex]

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

[tex]P(N|T)=\dfrac{P(N\cap T)}{P(T)}\\\\P(N|T)=\dfrac{P(N).P(T)}{P(T)}\\\\P(N|T)=P(N)=0.13[/tex]

Hence, our required probability is 0.13.

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