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Two dice are thrown simultaneously. The probability that sum is odd or less than $7$ or both, is
$\frac{2}{3}$
$\frac{1}{2}$
$\frac{3}{4}$
$\frac{1}{3}$
Solution
(c) Required probability
$ = P({\rm{less}}\,\,{\rm{than}}\,\,7) + P({\rm{odd}}) + P({\rm{both}}) – P(7 \cap {\rm{odd}})$
$ – P(7 \cap {\rm{both}}) – P({\rm{odd}} \cap {\rm{both}}) + P({\rm{odd}} \cap 7 \cap {\rm{both}})$
But $P({\rm{both}}) = P(7 \cap {\rm{odd}}) = P(7 \cap {\rm{both}}) = P(o{\rm{dd}} \cap {\rm{both}})$
$ = P({\rm{odd}} \cap 7 \cap {\rm{both}})$
Therefore required probability
$ = P({\rm{Less\, than 7)}} + P({\rm{odd) — }}P({\rm{7 }} \cap {\rm{odd}})$
$P({\rm{odd}}) = \frac{{18}}{{36}} = \frac{1}{2}$
$P({\rm{less}}\,\,{\rm{than}}\,\,7) = \frac{{15}}{{36}} = \frac{5}{{12}},\,\,P({\rm{both}}) = \frac{6}{{36}} = \frac{1}{6}$
Hence required probability $ = \frac{5}{{12}} + \frac{1}{2} – \frac{2}{{12}} = \frac{9}{{12}} = \frac{3}{4}.$
