# Six dice are thrown 729 times. How many times do you expect atleast three dice to show 5 or 6.

Probability of getting a 5 or 6 = 2/6 = 1/3

Hence p= 1/3 and q - 1-1/3 = 2/3, n = 6 and r = 3

Therefore probability of getting 5 or 6 in atleat 3 dice is 1- [P(X=0) + P(X=1)+P(X=2)]

= 1-.[(2/3)^6.(1/3)+6(2/3)^5(1/3)+15(2/3)^4(1/3)^2]

= 1 - [(16/9)(31/9)

= 1-426/729 = 233/729

Therefore for 729 trials . we can expect  729 x 233/729 =233 times