A basket contains 12 apples in which 3 are ro | vedclass

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(B) Total number of apples $= 12$. Number of rotten apples $= 3$. Number of good apples $= 9$.Total ways to draw $3$ apples from $12$ is ${}^{12}C_3 =

Vedclass · Accepted Answer

$\frac{48}{55}$

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(B) Total number of apples $= 12$. Number of rotten apples $= 3$. Number of good apples $= 9$.Total ways to draw $3$ apples from $12$ is ${}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.We need the probability of getting at most one rotten apple,which means $0$ or $1$ rotten apple.Case $1$: No rotten apple is drawn (all $3$ are good).Number of ways $= {}^{9}C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.Case $2$: Exactly $1$ rotten apple is drawn (and $2$ are good).Number of ways $= {}^{3}C_1 	imes {}^{9}C_2 = 3 	imes \frac{9 	imes 8}{2 	imes 1} = 3 	imes 36 = 108$.Total favorable ways $= 84 + 108 = 192$.Required probability $= \frac{192}{220} = \frac{48}{55}$.

$A$ basket contains $12$ apples in which $3$ are rotten. If $3$ apples are drawn at random simultaneously from it,then the probability of getting at most one rotten apple is

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