Suppose that Vase 1 contains 8 white balls and 2 green balls

Suppose that Vase 1 contains 8 white balls and 2 green balls. Then suppose that Vase 2 contains 3 white balls and 7 green balls. An experiment consists of tossing an unfair coin (probability of heads is 75%), then picking a ball from one of the vases. If the coin comes up heads, then you pick from Vase 1. If the coin comes up tails, then you pick from Vase 2 What is the probability you pick a green ball? P.S. i got 32.5% but I want to be sure. Show all steps please

Solution

Let A denote the event that the first vase is selected i.e. head is obtaibed in the first toss.

Given that P(A)= 0.75

Let B denote the event that the selected ball is green.

Then, P(B|A) = P[ Green ball is selected from the first vase] = 2 / (8+2) = 0.2

P(B|A^c)= P[Green ball is selected from second vase] = 7 / (7+3) = 0.7

Required to find P(B).

By the theorem of total probability,

P(B) = P(B|A)*P(A) + P(B|A^c)*P(A^c)

        = 0.2 * 0.75 + 0.7 * (1 - 0.75)     [P(A^c) = 1 - P(A) ]

        = 0.15 + 0.175

        =0.325

Hence the probability of picking green ball is 32.5 % .