Question

A television​ network, Network​ A, is scheduling its fall lineup of shows. For the Tuesday night 8 p.m.​ slot, Network A has selected its top show. If its rival​ network, Network​ B, schedules its top show during the same time​ slot, Network A estimates that it will get 1.1 million viewers.​ However, if Network B schedules a different show during that time​ slot, Network A estimates that it will get 1.9 million viewers. Network A believes that the probability that Network B will air their top show is 0.7 and the probability that Network B will air another show is 0.3. Determine the expected number of viewers for Network​ A's top show.

Answer 1

Answer:

1,280,000 (1.28 million.)

Step-by-step explanation:

If Network B schedules its top show (with a probability of 0.7), Network A will get 1.1 million viewers.

If Network B schedules a different show during that time​ slot, (with a probability of 0.3), Network A will get 1.9 million viewers.

Therefore, the probability distribution table of number of viewers of Network A is:

$\left|\begin{array}{c|c|c} Number of Viewers, x&1.1 million& 1.7 million\\P(x)&0.7&0.3\end{array}\right|$

Therefore, the expected number of viewers for Network​ A's top show

= (1100000 X 0.7) + (1700000 X 0.3)

=1,280,000

The expected number of viewers for Network​ A's top show is 1.28 million.