A bag contains 6 red, 2 white and 8 blue ball | vedclass

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Question

(A) Total number of balls $= 6 + 2 + 8 = 16$.Total ways to draw $3$ balls $= ^{16}C_3 = \frac{16 	imes 15 	imes 14}{3 	imes 2 	imes 1} = 560$.$A$.

Vedclass · Accepted Answer

$A-III, B-I, C-IV, D-II$

Vedclass · Answer

(A) Total number of balls $= 6 + 2 + 8 = 16$.Total ways to draw $3$ balls $= ^{16}C_3 = \frac{16 	imes 15 	imes 14}{3 	imes 2 	imes 1} = 560$.$A$. Probability that none of the balls is white (i.e.,all $3$ are from $6$ red and $8$ blue balls):$P(A) = \frac{^{14}C_3}{^{16}C_3} = \frac{14 	imes 13 	imes 12}{16 	imes 15 	imes 14} = \frac{13}{20} = III$.$B$. Probability of getting $2$ white and $1$ blue ball:$P(B) = \frac{^{2}C_2 	imes ^{8}C_1}{^{16}C_3} = \frac{1 	imes 8}{560} = \frac{1}{70} = I$.$C$. Probability of getting $2$ blue and $1$ white ball:$P(C) = \frac{^{8}C_2 	imes ^{2}C_1}{^{16}C_3} = \frac{28 	imes 2}{560} = \frac{56}{560} = \frac{1}{10} = IV$.$D$. Probability of getting $1$ red, $1$ white and $1$ blue ball:$P(D) = \frac{^{6}C_1 	imes ^{2}C_1 	imes ^{8}C_1}{^{16}C_3} = \frac{6 	imes 2 	imes 8}{560} = \frac{96}{560} = \frac{6}{35} = II$.Thus, the correct match is $A-III, B-I, C-IV, D-II$.

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$A$. Probability that none of the balls is white	$I$. $\frac{1}{70}$
$B$. Probability of getting $2$ white and $1$ blue ball	$II$. $\frac{6}{35}$
$C$. Probability of getting $2$ blue and $1$ white ball	$III$. $\frac{13}{20}$
$D$. Probability of getting $1$ red, $1$ white and $1$ blue ball	$IV$. $\frac{1}{10}$