A child rolls a 6-sided die 6 times. What is the probability of the child rolling no more than three twos?
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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so we are looking for the probability that no two’s show up plus 1 two’s shows up plus 2 two’s show up and 3 two’s show up
the probability of a two shows up is 1/6
no two
N N N N N N ==> (5/6)^6
1 two
2 N N N N N ==> 6(1/6) (5/6)^5
2 two’s
2 2 N N N N ==> 15 (1/6)^2 (5/6)^4
3 two’s
2 2 2 N N N ==> 20 (1/6)^3 (5/6)^3
(5/6)^6 + 6(1/6) (5/6)^5 + 15(1/6)^2 (5/6)^4 + 20(1/6)^3 (5/6)^3
= .9913
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is it 625/46656
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