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

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14.Probability

hard

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

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

$\frac{8}{{21}}$

$\frac{{32}}{{12}}$

None of these

Solution

(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} \times {}^{32}{C_1} + {}^{32}{C_1} \times {}^{32}{C_2}.$
Hence the required probability $ = \frac{{2\left( {{}^{32}{C_1}.{}^{32}{C_2}} \right)}}{{{}^{64}{C_3}}} = \frac{{16}}{{21}}.$

Standard 11

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