Discussion

Explanation:

Let us first calculate P (1 ball is blue and 1 red) = P(1st blue and 2nd red) + P (1st red and 2nd blue).

P(1st blue) = 515

Now, there are 14 balls remaining.

P(2nd red) = 614

P(1st blue and 2nd red) = 515×614

Similarly, P (1st red and 2nd blue) = 615×514

P (1 ball is blue and 1 red) = 515×614 + 615×514 = 60/210.

Similarly

P (1 ball is blue and 1 black) = 515×414 + 415×514 = 40/210

P (1 red is blue and 1 black) = 615×414 + 415×614 = 48/210

∴ P (both balls are of different color) = P (1 ball is blue and 1 red) + P (1 ball is blue and 1 black) + P (1 red is blue and 1 black) 

⇒ ∴ P (both balls are of different color) = 60210+40210+48210 = 148210 = 74105

Hence, option (b).

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