Question

one bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls . we first toss a fair coin .if it is head , one ball is drawn from the first bag and placed in the second bag.if it is a tail , one ball is drawn from the second bag and placed in the first bag.let the event A be that a ball is now drawn from the second bag is black .find out p(A)

Answer 1

P(A) = n(E)/n(S)

Where P(A) is the probability of an event A

n(E) is the number of favorable outcomes

n(S) is the total number of events in the sample space.

probability of getting the second bag(tail) = 1/2

probability of getting black ball from second bag = 5/8

P(A) = (1/2)*(5/8) = 5/16 = 0.3125

The probability of event A is 0.3125.