Question

An advertising executive studying television viewing habits of married men and women during​ prime-time hours has determined that during prime​ time, husbands are watching television 80 % of the time. When the husband is watching​ television, 50 % of the time the wife is also watching. When the husband is not watching​ television, 10 % of the time the wife is watching television. a. Find the probability that if the wife is watching​ television, the husband is also watching television. b. Find the probability that the wife is watching television in prime time.

Answer 1

Answer:

a) 0.9523

b) 0.42

Step-by-step explanation:

Given:

husbands are watching television at prime time, P(H) = 0.80

husbands not watching television at prime time, P(H')= 1 - 0.80 = 0.20

When the husband is watching​ television, wife is also watching,

P( W | H ) = 0.50

When the husband is not watching​ television, wife is watching

⇒ P( W | H') = 0.10

Now,

a) The probability that wife is watching TV

⇒ P(W) = P( W | H')P(H') + P( W | H )P(H)

= (0.10 × 0.20) + (0.50 × 0.80)

= 0.02 + 0.4

= 0.42

By the Baye's theorem,  

P( wife is watching​ television, when the husband is also watching tv)

⇒ P( H | W ) =$\frac{P(W|H)P(H)}{P(W)}$

= $\frac{0.50\cdot 0.80}{0.42}$

= 0.9523

b) P( wife is watching TV in prime time)

P(W) = P( W | H')P(H') + P( W | H )P(H)

= ( 0.10 × 0.20 ) + ( 0.50 × 0.80 )

= 0.02 + 0.4

= 0.42