(b) Let E
1 , E
2 and A be the events defined as follows:
E
1 = red ball is transferred from bag P to bag Q
E
2 = blue ball is transferred from bag P to bag Q
A = the ball drawn from bag Q is blue
As the bag P contains 6 red and 4 blue balls,
P(E
1) =
106=53 and P(E
2) =
104=52Note that E
1 and E
2 are mutually exclusive and exhaustive events.
When E
1 has occurred i.e., a red ball has already been transferred from bag P to Q, then bag Q will contain 6 red and 6 blue balls, So, P(A|E
1) =
126=61 When E
2 has occurred i.e., a blue ball has already been transferred from bagP to Q, then bag Q will contain 5 red and 7 blue balls, So,P(A|E
2) =
127 By using law of total probability, we get P(A) = P(E
1) P(A|E
1) + P(E
2) P(A|E
2)
=
53×21+52×127 =
158