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

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Question

(a) We have the following three pattern :

$(i)$ Red, white $P(A) = \frac{{3 	imes 4}}{{{}^{12}{C_2}}}$

$(ii)$ Red, blue $P(B) = \frac{{3 	imes

Vedclass · Accepted Answer

$\frac{{47}}{{66}}$

Vedclass · Answer

(a) We have the following three pattern :

$(i)$ Red, white $P(A) = \frac{{3 	imes 4}}{{{}^{12}{C_2}}}$

$(ii)$ Red, blue $P(B) = \frac{{3 	imes 5}}{{{}^{12}{C_2}}}$

$(iii)$ Blue, white $P(C) = \frac{{4 	imes 5}}{{{}^{12}{C_2}}}$

Since all these cases are exclusive, so the required probability

$ = \frac{{(12 + 15 + 20)}}{{{}^{12}{C_2}}} = \frac{{(47 	imes 2)}}{{(12 	imes 11)}} = \frac{{47}}{{66}}.$

A bag contains $3$ red, $4$ white and $5$ blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is

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