A box contains 8 red and 7 black balls. If tw | vedclass

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(C) Total number of balls = $8 + 7 = 15$. Number of ways to select $2$ balls from $15$ is $n(S) = \binom{15}{2} = \frac{15 	imes 14}{2} = 105$. Let $

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$\frac{7}{15}$

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(C) Total number of balls = $8 + 7 = 15$. Number of ways to select $2$ balls from $15$ is $n(S) = \binom{15}{2} = \frac{15 	imes 14}{2} = 105$. Let $E$ be the event that both balls are of the same color. This means either both are red or both are black. Number of ways to select $2$ red balls = $\binom{8}{2} = \frac{8 	imes 7}{2} = 28$. Number of ways to select $2$ black balls = $\binom{7}{2} = \frac{7 	imes 6}{2} = 21$. Number of favorable outcomes $n(E) = 28 + 21 = 49$. Probability $P(E) = \frac{n(E)}{n(S)} = \frac{49}{105} = \frac{7}{15}$.

$A$ box contains $8$ red and $7$ black balls. If two balls are drawn at random,the probability that both balls are of the same color is ...... .

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