Discussion

Explanation:

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

P (1^st blue) = 5/15

Now, this ball is again replaced, hence we again have a total of 15 balls to pick from.

P (2^nd red) = 6/15

P (1^st blue and 2^nd red) = 5$\frac{5}{15}\times \frac{6}{15}$

Similarly, P (1^st red and 2^nd blue) = $\frac{6}{15}\times \frac{5}{15}$

P (1 ball is blue and 1 red) = $\frac{5}{15}\times \frac{6}{15}$ + $\frac{6}{15}\times \frac{5}{15}$ = $\frac{60}{225}$.

Similarly

P (1 ball is blue and 1 black) = $\frac{5}{15}\times \frac{4}{15}$ + $\frac{4}{15}\times \frac{5}{15}$ = $\frac{40}{225}$

P (1 red is blue and 1 black) = $\frac{6}{15}\times \frac{4}{15}$ + $\frac{4}{15}\times \frac{6}{15}$ = $\frac{48}{225}$

∴ 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) = $\frac{60}{225}+\frac{40}{225}+\frac{48}{225}$ = $\frac{148}{225}$

Hence, option (c).

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