Q. 124.8( 4 Votes )

One bag contains

Answer :

We have 2 possibilities –


1. red transferred and red drawn


2. black transferred and red drawn


Total balls in first bag = 8


Total balls in bag 2 = 10


Case – 1:


Initially, red ball is transferred from the first bag.


So,the probability of transferring a red ball from first bag


Now, the second bag has 7 red and 4 black balls,


So,probability of drawing a red ball from second bag


So, required probability from case 1



Case 2:


Initially, black ball is transferred from first bag.


So, probability of transferring a black ball from first bag


Now, second bag has 6 red and 5 black balls,


So,probability of drawing a red ball from second bag


So,required probability from case 2



Also, Case 1 and Case 2 are independent,


So,required probability



OR


P(A) = 0.6,


P(B) = 0.5,


P(A|B) = 0.3




P(AB) = 0.3(P(B))


P(AB) = (0.3) (0.5)


P(AB) = 0.15


Now, P(AB) = P(A) + P(B) – P(AB)


P(AB) = 0.6 + 0.5 – 0.15


= 0.95


Therefore, P(AB) = 0.95


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