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(AꓵB) = 0.3(P(B))

⇒ P(AꓵB) = (0.3) (0.5)

⇒ P(AꓵB) = 0.15

Now, P(AꓴB) = P(A) + P(B) – P(AꓵB)

⇒ P(AꓴB) = 0.6 + 0.5 – 0.15

= 0.95

Therefore, P(AꓴB) = 0.95

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