Two dice are thrown simultaneously. The proba | vedclass

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Question

(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}}) -

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(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}.$

Two dice are thrown simultaneously. The probability that sum is odd or less than $7$ or both, is

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