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.

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 ) =

=

= 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

stops on an even number.

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)