Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.

Explanation

Solution

P(Sum 6)=`frac{5}{36} `

`(1,5),(5,1),(2,4),(4,2),(3,3) `

P(Sum 7)=`frac{6}{36}=frac{1}{6} `

`(1,6),(6,1),(2,5),(5,2),(3,4),(4,3) `

`therefore` P(A wins)`=frac{5}{36}+frac{31}{36} times frac{5}{6} times frac{5}{36}+frac{31}{36} times frac{5}{6} times frac{31}{36} times frac{5}{6} times frac{5}{36} `

=`frac{frac{5}{36}}{1-frac{31}{36} times frac{5}{6}}=frac{30}{61}`