# One bag contains

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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