ProbabilityHard
Question
Urn A contains 6 red & 4 black balls and urn B contains 4 red & 6 black balls. One ball is drawn at random from urn A & placed in urn B. Then one ball is drawn at random from urn B & placed in urn A. If one ball is now drawn at random from urn A, the probability that it is red is
Options
A.19/55
B.32/55
C.41/55
D.9/55
Solution
Let A1 → Ball drawn from urn A is red and ball returned is also red, P(A1) = 
B1 → Ball drawn from urn A is red but ball returned to it is black, P(B1) =
C1 → Ball drawn from urn A is black and ball of same colour is returned, P(C1) =
D1 → Ball drawn from urn A is black and ball returned is red, P(D1) =
Required probability P(R) = P(A1) × P
+ P(B1) × P
+ P(C1) × P
+ P(D1) × P
= P(A1) ×
+ P(B1) ×
+ P(C1) ×
+ P(D1) × 
B1 → Ball drawn from urn A is red but ball returned to it is black, P(B1) =
C1 → Ball drawn from urn A is black and ball of same colour is returned, P(C1) =
D1 → Ball drawn from urn A is black and ball returned is red, P(D1) =
Required probability P(R) = P(A1) × P
= P(A1) ×
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