A box contains 15 black and 20 white balls. W | vedclass

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(A) Total number of balls = $15 + 20 = 35$.Number of ways to select $2$ balls from $35$ is $^{35}C_2 = \frac{35 	imes 34}{2} = 595$.'At most one whit

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$\frac{81}{119}$

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(A) Total number of balls = $15 + 20 = 35$.Number of ways to select $2$ balls from $35$ is $^{35}C_2 = \frac{35 	imes 34}{2} = 595$.'At most one white ball' means either $0$ white balls (both black) or $1$ white ball (one black and one white).Case $1$: Both balls are black. Number of ways = $^{15}C_2 = \frac{15 	imes 14}{2} = 105$.Case $2$: One white and one black ball. Number of ways = $^{20}C_1 	imes ^{15}C_1 = 20 	imes 15 = 300$.Total favorable outcomes = $105 + 300 = 405$.Probability = $\frac{405}{595} = \frac{81}{119}$.

$A$ box contains $15$ black and $20$ white balls. What is the probability that when $2$ balls are selected,at most one is white?

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