A bag contains 4 white, 5 red and 6 black bal | vedclass

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Question

(a) Required probability $=P(WR) + P(WB) $$= \frac{{^4{C_1}{(^5}{C_1}{ + ^6}{C_1})}}{{^{15}{C_2}}}$

$ = \frac{{{}^4{C_1}\,.\,11}}{{{}^{15}{C_2}}} =

Vedclass · Accepted Answer

$\frac{{44}}{{105}}$

Vedclass · Answer

(a) Required probability $=P(WR) + P(WB) $$= \frac{{^4{C_1}{(^5}{C_1}{ + ^6}{C_1})}}{{^{15}{C_2}}}$

$ = \frac{{{}^4{C_1}\,.\,11}}{{{}^{15}{C_2}}} = \frac{{4\,.\,11\,.\,2}}{{15\,.\,14}} = \frac{{44}}{{105}}.$

A bag contains $4$ white, $5$ red and $6$ black balls. If two balls are drawn at random, then the probability that one of them is white is

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