Question

4. A bag contains 1 red, 3 green, and 5 yellow balls. A sample of four balls is picked. Let G be the number of green balls in the sample. Let Y be the number of yellow balls in the sample. Find the conditional probability mass function of G given Y 2 assuming the sample is picked with replacement

Answer 1

The following joint PMFs can be obtained (Use trinomial distribution).

4! 2) = 2121 (3/9) (1/9)2 (5/9)2 = 150/6561 P(G 0, Y 4! P(G = 1,Y = 2) = 1121 (3/9) (1/9) (5/9) = 900/6561 4! 2) = 2121 (3/9)2(1/9)° (5/9)2 = 1350/6561 P(G 2,Y

Now, the probability of 2 green balls in a draw of 4 balls with replacement is,

4! P(Y 2) (5/9)2 (4/9)2 2400/6561 =

The conditional PMF of given
YB2
are

01Y 2) = P(G = 0,Y = 2 P(G P(Y 2 150/6561 P(G 01Y = 2) = 2400/6561 P(G 0|Y 2) = 1/16 P(G = 1|Y = 2) = P(G = 1, Y = 2) P(Y 2 900/6561 P(G = 1|Y = 2) = 2400/6561 P(G = 1|Y = 2) = 3/8 P(G = 2|Y = 2) = P(G = 1, Y = 2) 1350/6561 P(Y 2 P(G = 2|Y = 2) = 2400/6561 P(G 2|Y 2) 9/16

It can be see that $\fn_jvn \sum P(G=g|Y=2)=1$ . Property of conditional PMF.