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television network is about to telecast a new television show. Before a show is launched, the network airs a pilot episode and receives a report assessing favorable or unfavorable viewer response. In the past, 60% of the network's shows have received a favorable response from viewers, and 40% have received an unfavorable response. If 50% of the network’s shows have received a favorable response and have been successful, and 30% of the network’s shows have received an unfavorable response and have been successful, what is the probability that this new show will be successful if it receives a favorable response?

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merlynthewhizz
We have:

P(Favorable) = 0.6 and from this we have P(Successful) = 0.5 and P(unsuccesful) = 0.5

P(Unfavorable) = 0.4 and from this, we have P(Successful) = 0.3 and P(Unsuccesful) = 0.7

P(Successful) = (0.6×0.5) + (0.4×0.3) = 0.3 + 0.12 = 0.42

The question is a conditional probability: what is the probability of a program being successful GIVEN a program is favourable.

P(Successful | Favourable) = P(Successful∩Favourable) / P(Favourable)
P(S | F) = 0.42/0.6 = 0.7
D4B

Answer:

The probability that this new show will be successful if it receives a favorable response is 0.833

Step-by-step explanation:

We are going to solve this problem using conditional probability. From the question lets state some of the conditions.

Let X be the event that the TV show is successful, so the probability that the TV show is successful is P(X) = 0.5

Let X' be the event that the TV show is unsuccessful, so the probability that the TV show is unsuccessful is P(X') = 0.5

Let Y be the event that there was a favorable response, so the probability that the show had a favorable response P(Y) = 0.6

Let Y' be the event that there was an  unfavorable response, so the probability that the show had an unfavorable response P(Y') = 0.4

If 50% of the network’s shows have received a favorable response and have been successful then,

P(X∩Y) = 0.5

and 30% of the network’s shows have received an unfavorable response and have been successful then,

P(X∩Y') = 0.3

The probability that this new show will be successful if it receives a favorable response will be

P(X/Y) =[tex]\frac{P(XnY)\}{P(Y)}[/tex] =[tex]\frac{0.5}{0.6}[/tex]

P(X/Y) = 0.833

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