# roll a die 10 times. find chance of exactly 4 twos?

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• Use the binomial probability formula .

P( exactly k events in n trials) = (n k) p^(k) (1-p)^(n-k)

where (n k ) = nCk = n! / ( (n-k)! k!)

n = number of trials

k = exact number of successes

p = probability of success in one trial

so

P(exactly 4 twos in 10 trials) = ( 10 4) (1/6)^4 (5/6)^6

P(exactly 4) = (10 4) (5^6/6^10)

(10 4) = 10! / (6! *4!) = 10*9*8*7 / 4*3*2*1

(10 4) = 10*9*7/3 = 10*3*7 = 210

p(exactly 4) = 210 * 5^6/6^10

P(exactly 4 ) = 3281250 /60466176

P(exactly 4 ) = 546875 / 10077696

P(exactly 4 ) = approx. 0.054265876

so about a 5.43 % chance so not very probable

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• The number of 2s has a binomial distribution.

p = probability of success on any single trial = 1/6

n = number of trials = 10

r = number of successes = 4

required probability = C(n, r) p^r (1 - p)^(n - r)

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• 50/50

either it happens or it doesn't happen

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