A bag contains 3 red, 5 black and 7 blue ball | vedclass

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Question

(A) Total number of balls = $3 + 5 + 7 = 15$. Number of ways to draw $3$ balls from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	ime

Vedclass · Accepted Answer

$\frac{29}{65}$

Vedclass · Answer

(A) Total number of balls = $3 + 5 + 7 = 15$. Number of ways to draw $3$ balls from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$. We need to find the probability of getting at least two blue balls. This can happen in two cases: Case $1$: Exactly $2$ blue balls and $1$ non-blue ball. Number of ways = $^7C_2 	imes ^8C_1 = 21 	imes 8 = 168$. Case $2$: Exactly $3$ blue balls. Number of ways = $^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$. Total favorable outcomes = $168 + 35 = 203$. Probability = $\frac{203}{455} = \frac{29 	imes 7}{65 	imes 7} = \frac{29}{65}$.

$A$ bag contains $3$ red, $5$ black and $7$ blue balls. If three balls are drawn at random simultaneously from the bag, then the probability of getting at least two blue balls is

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