Aman verma wrote:Q : Urn A contains six red and four black balls and Urn B contains four red and six black balls.One ball is drawn at random from Urn A and placed in Urn B.Then one ball is transferd at random from Urn B to Urn A .If one ball is now drawn at random from Urn A ,find the probability that it is red :
a)32/65
b32/55
c)23/55
d)56/65
e)31/65
Case 1: One red ball is transferred to the second urn
Probabilty of that happening = 6/10 = 3/5
One red ball is transferred to the first urn from second
Probabilty of that happening = 5/11 = 5/11
One red ball is selected from urn A after this = 6/10
Probability that all of them happening together (all of the events are independent) = 3/5*5/11*3/5 = 9/55
Case 2: Red ball is first transferred, black ball is transferred back and red ball is chosen
Probability 6/10*6/11*5/10 = 9/55
Case 3: Black ball is first transferred, red ball is transferred back and red ball is chosen
Probability = 4/10*4/11*5/10 = 2/5*1/2*4/11 =4/55
Case 4: Black ball is first transferred , black ball is transferred back and red ball is chosen
4/10*7/11*6/10 =2/5*3/5*7/11= 42/5* 1/55 = 8.4/55
I am messing up calculation somewhere and getting 30.4/55
Always borrow money from a pessimist, he doesn't expect to be paid back.
