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fgiga [73]
3 years ago
9

Suppose each of 12 players rolls a pair of dice 3 times. Find the probability that at least 4 of the players will roll doubles a

t least once. (Answer correct to four decimal places.)
Mathematics
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

Our answer is 0.8172

Step-by-step explanation:

P(doubles on a single roll of pair of dice) =(6/36) =1/6

therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)

=1-(1-1/6)3 =91/216

for 12 players this follows binomial distribution with parameter n=12 and p=91/216

probability that at least 4 of the players will get “doubles” at least once =P(X>=4)

=1-(P(X<=3)

=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)

=1-0.1828

=0.8172

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