Question

Find the minimum number of tosses of a pair of dice so that the probability of getting the sum of the digits on the dice equal to $7$ on atleast one toss, is greater than $0.95$. (Given ${\log }_{10}2=\mathrm{0.3010.}{\log }_{10}2=0.477$).Minimum number of tosses is given as $20-k$.Find the value of $k$.

Answer 1

$n\left(S\right)=36$
Let E be the event of getting the sum of digits on the dice equal to 7, then $n\left(E\right)=6$
$P\left(E\right)=\frac{6}{36}=\frac{1}{6}=p$
then $P\left(E^{\prime }\right)=q=\frac{5}{6}$
probability of not throwing the sum 7 in first m trails $=q^m$
Therefore P(at least one 7 in m throw) $=1-q^m=1-{\left(\frac{5}{6}\right)}^m$
According to the question
$1-{\left(\frac{5}{6}\right)}^m>0.95\Rightarrow {\left(\frac{5}{6}\right)}^m>0.05\Rightarrow m({\log }_{10}5-{\log }_{10}6)<{\log }_{10}1-{\log }_{10}20\therefore m>16.44$
Hence, the least number of trails $=17$