Six dice are rolled .calling a 5 or 6 a success .Find the probability of getting at least 2 failure

Answer 1

p = Prob(success) = Prob ( 5 or 6) = 2/6 = 1/3

So q = Prob(failure) = Prob( 1,2,3,4) = 4/6 = 2/3 = 1 - 1/3 = 1 - Prob(success)

X is BInomial ( p=1/3, N=6, k=4,3,2,1,0) <--- Prob of at least 2 failures = Prob of 4 or less successes

Prob ( X <= 4) = 1 - Prob(X>4) = 1 - [Prob(X=5) + Prob(X=6)] = 1 - Prob(X=5) - Prob(X=6)

Prob(X=5) = (6 choose 5)* (1/3)^5 * (2/3)

= 6 * 1/243 * 2/3 =

4/243

Prob(X=6) = (6 choose 6) p^6 q^0 = 1 * (1/3)^6 * 1 = 1/729

1 - 4/243 - 1/729 = (729 - 12 - 1) / 729

= 716/729

= 98.2167...%