dice probability?
A child rolls a 6-sided die 6 times. What is the probability of the child rolling no more than three twos?
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Use the binomial formula:
(p + q)^6 =
p^6 + 6p^5q + 15p^4q^2 + 20p^3q^3 + 15p^2q^4 + 6pq^5 + q^5
p = 1/6 or .16666
q = 5/6 or .83333
No more than 3 twos is expressed as: P(x>3)
P(x>3) = 20p^3q^3 + 15p^2q^4 + 6pq^5 + q^5
P(x>3) = .053584 + .200939 + .401878 + .334898
P(x>3) = .991298 or .9913
Good luck in your studies,
~ Mitch ~
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