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Question

(a) Three squares can be chosen out of $64$ squares in ${}^{64}{C_3}$ ways. Two squares of one colour and one another colour can be chosen in two mutu

Vedclass · Accepted Answer

$\frac{{16}}{{21}}$

Vedclass · Answer

(a) Three squares can be chosen out of $64$ squares in ${}^{64}{C_3}$ ways. Two squares of one colour and one another colour can be chosen in two mutually exclusive ways :
$(i)$ Two white and one black and $(ii)$ Two black and one white. Thus the favourable number of cases
$ = {}^{32}{C_2} 	imes {}^{32}{C_1} + {}^{32}{C_1} 	imes {}^{32}{C_2}.$
Hence the required probability $ = \frac{{2\left( {{}^{32}{C_1}.{}^{32}{C_2}} ight)}}{{{}^{64}{C_3}}} = \frac{{16}}{{21}}.$

Three squares of a chess board are chosen at random, the probability that two are of one colour and one of another is

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