A bag contains 4 red, 3 white, and 5 blue bal | vedclass

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Question

(A) Total number of balls = $4 + 3 + 5 = 12$.Total ways to draw $3$ balls = $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.$(A)

Vedclass · Accepted Answer

Vedclass · Answer

(A) Total number of balls = $4 + 3 + 5 = 12$.Total ways to draw $3$ balls = $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.$(A)$ Probability of getting $1$ red, $1$ white and $1$ blue ball:Ways = $^4C_1 	imes ^3C_1 	imes ^5C_1 = 4 	imes 3 	imes 5 = 60$.Probability = $\frac{60}{220} = \frac{3}{11}$ (Matches $(iv)$).$(B)$ Probability of getting $2$ white and $1$ blue ball:Ways = $^3C_2 	imes ^5C_1 = 3 	imes 5 = 15$.Probability = $\frac{15}{220} = \frac{3}{44}$ (Matches $(i)$).$(C)$ Probability of getting $2$ red and $1$ white ball:Ways = $^4C_2 	imes ^3C_1 = 6 	imes 3 = 18$.Probability = $\frac{18}{220} = \frac{9}{110}$ (Matches $(v)$).$(D)$ Probability that none of the balls is white (i.e.,all $3$ are from red and blue balls, total $4+5=9$):Ways = $^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.Probability = $\frac{84}{220} = \frac{21}{55}$ (Matches $(ii)$).Therefore, the correct matching is: $A ightarrow (iv), B ightarrow (i), C ightarrow (v), D ightarrow (ii)$.

List-$I$	List-$II$
$A$. Probability of getting $1$ red, $1$ white and $1$ blue ball	$(i)$ $\frac{3}{44}$
$B$. Probability of getting $2$ white and $1$ blue ball	$(ii)$ $\frac{21}{55}$
$C$. Probability of getting $2$ red and $1$ white ball	$(iii)$ $\frac{38}{55}$
$D$. Probability that none of the balls is white	$(iv)$ $\frac{3}{11}$
	$(v)$ $\frac{9}{110}$

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