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 $0.95$ $({\log }_{10}2=0.3010;{\log }_{10}3=0.4771;{\log }_{10}5=0.6990$)

Answer 1

Probability of getting sum $7$ is $\frac{1}{6}$
Probability of not getting sum equal to $7$ is $\frac{5}{6}$
Therefore required probability is
$\frac{1}{6}+\frac{5}{6}.\frac{1}{6}+{\left(\frac{5}{6}\right)}^2.\frac{1}{6}+...+{\left(\frac{5}{6}\right)}^{n-1}.\frac{1}{6}=\frac{1}{6}\frac{1-{\left(\frac{5}{6}\right)}^n}{1-\frac{5}{6}}=1-{\left(\frac{5}{6}\right)}^n$
And this exceeds $0.95$
$1-{\left(\frac{5}{6}\right)}^n>\frac{95}{100}\Rightarrow {\left(\frac{5}{6}\right)}^n<\frac{5}{100}\Rightarrow n>16.45\Rightarrow n=17$