A box contains 5 red,4 green and 3 black ball | vedclass

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Question

(B) Total number of balls $= 5 + 4 + 3 = 12$.The total number of ways to draw $3$ balls from $12$ is given by $n(S) = {}^{12}C_{3} = \frac{12 	imes 1

Vedclass · Accepted Answer

$\frac{41}{44}$

Vedclass · Answer

(B) Total number of balls $= 5 + 4 + 3 = 12$.The total number of ways to draw $3$ balls from $12$ is given by $n(S) = {}^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.Let $E$ be the event that all $3$ balls are of the same colour.This can happen if all $3$ are red,all $3$ are green,or all $3$ are black.Number of ways to choose $3$ red balls $= {}^{5}C_{3} = 10$.Number of ways to choose $3$ green balls $= {}^{4}C_{3} = 4$.Number of ways to choose $3$ black balls $= {}^{3}C_{3} = 1$.Total ways for same colour $= 10 + 4 + 1 = 15$.Probability that all balls are of the same colour $P(E) = \frac{15}{220} = \frac{3}{44}$.The probability that they are not of the same colour $= 1 - P(E) = 1 - \frac{3}{44} = \frac{41}{44}$.

$A$ box contains $5$ red,$4$ green and $3$ black balls. $3$ balls are drawn at random. What is the probability that they are not of the same colour?

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