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 at least once. (Answer correct to four decimal places.)

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