6 dice are thrown 729 times. How many times do you expect atleast three dice to show a five or six?
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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
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
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