
Dave S. answered 03/19/16
Tutor
4.8
(133)
Math, Science & Engineering Tutor
Since one ball is drawn and one is replaced there will always be 8 balls in the urn.
a) after drawing Y1, the urn will contain 4 yellow and 4 green. Pr(Y2|Y1) = 4/8 or 1/2
b) after drawing G1, the urn will contain 6 yellow and 2 green. Pr(Y2|G1) = 6/8 or 3/4
c) after drawing Y1, and then Y2, the urn will contain 3 yellow and 5 green. Pr(Y3|Y2uY1) = 3/8
d) after drawing Y1, and then G2, the urn will contain 5 yellow and 3 green. Pr(Y3|G2uY1) = 5/8
Note if Y2 means that the second ball drawn is yellow; then Pr(Y2|Y1uG2) = 0 since the second ball drawn was a green.