A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing. Compute P ( T ) . Compute P ( T | F ) . Are T and F independent

P (T) = 1/4

P ( T | F )    = 1/2  = P(F)

The events are not independent.

Step-by-step explanation:

Let F the event of picking the white ball first

P (F)= 1/2   ( picking the white ball first)

Let T be the event of getting the white ball twice,

P (T) = P( getting white ball) * P( getting white ball)

          =( 1/2)*(1/2)

           = 1/4

Here P(T∩F) = P(T) because the probability of getting the white balls is the same as probability of getting the white ball first  both the times.

P ( T | F ) = P (T∩F)/ P(F)

                = (1/4)/ (1/2)

                  = (1/2)

                    = 1/2  = P(F)

For the events to be independent the conditional probability P ( T | F )  must be equal to P(T).

Hence the events are not independent.

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

answer: im sorry i dont get it

step-by-step explanation: