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An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2, 3, 4,.......,12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$, is
$0.24$
$0.244$
$0.024$
None of these
Solution
(b) Required probability $=$ probability that either the number is $7$ or the number is $8$.
$i.e.,$ Required Probability $ = {P_7} + {P_8}$
Now ${P_7} = \frac{1}{2}.\frac{1}{{11}} + \frac{1}{2}.\frac{6}{{36}} = \frac{1}{2}\left( {\frac{1}{{11}} + \frac{1}{6}} \right)$
${P_8} = \frac{1}{2}.\frac{1}{{11}} + \frac{1}{2}.\frac{5}{{36}} = \frac{1}{2}\left( {\frac{1}{{11}} + \frac{5}{{36}}} \right)$
$\therefore \,\,\,P = \frac{1}{2}\left( {\frac{2}{{11}} + \frac{{11}}{{36}}} \right) = 0.244.$
