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

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.