Find the probability of throwing a sum of 8 at least ,9times in 10 throws of a pair of dice

Answer 1

Way to roll the 8:

2 and 6

3 and 5

4 and 4

5 and 3

6 and 2

So the probability of rolling the 8 is 5/36

Binomial p= 5/36, N=10 trials, k=9,10

Prob ( X>=9) = Prob(X=9) + Prob(x=10)

Prob(x=9) =

(10 choose 9) (5/36)^9 ( 31/36) = 1.6560243762... x 10^-7 which is essentially zero

Prob(x-10) =

(10 choose 10) (5/36)^10 ( 31/36)^0 = 1 * (5/36)^10 * 1 = 2.67100... x 10^-9 which is essentially zero

1.6827644467838989588741423304308 x 10^( -7)