Find the probability of throwing at teast 3 sixes in 5 throws of a dia.

Explanation

In one throw of a die p(getting a 60=`1/6`
p(getting a non 6 )=`5/6`
In 5 throws of a die
p(atleast 3 sixes)=p(3 sixer)+p(4 sixer)+p(5 sixer)
= `"^5C_3(1/6)^3(5/6)^2+^5C_4(1/6)^4(5/6)+("^5C_5)(1/6)^5=250/6^5+25/6^5+1/6^5=276/6^5`