A box contains 6 bottles of V_1 drink,3 bottl | vedclass

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(D) Total number of bottles = $6 + 3 + 4 = 13$. Number of ways to choose $3$ bottles out of $13$ is $^{13}C_3 = \frac{13 	imes 12 	imes 11}{3 	imes

Vedclass · Accepted Answer

$\frac{261}{286}$

Vedclass · Answer

(D) Total number of bottles = $6 + 3 + 4 = 13$. Number of ways to choose $3$ bottles out of $13$ is $^{13}C_3 = \frac{13 	imes 12 	imes 11}{3 	imes 2 	imes 1} = 286$. Let $E$ be the event that all three bottles are of the same variety. $E$ occurs if we choose $3$ bottles of $V_1$,or $3$ bottles of $V_2$,or $3$ bottles of $V_3$. Number of ways to choose $3$ bottles of the same variety = $^6C_3 + ^3C_3 + ^4C_3$. $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. $^3C_3 = 1$. $^4C_3 = 4$. Total favorable ways for $E = 20 + 1 + 4 = 25$. $P(E) = \frac{25}{286}$. The probability that the three bottles are not of the same variety is $1 - P(E) = 1 - \frac{25}{286} = \frac{286 - 25}{286} = \frac{261}{286}$.

$A$ box contains $6$ bottles of $V_1$ drink,$3$ bottles of $V_2$ drink and $4$ bottles of $V_3$ drink. If three bottles are drawn at random,then the probability that the three are not of the same variety is

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