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 at least one toss is greater than $0.95$(${\log }_{10}2=0.3010;{\log }_{10}3=0.4771$)

• A. $17$
• B. $18$
• C. $19$
• D. $20$

Answer 1

The correct option is A $17$

$n\left(S\right)=36$
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 = $17$