A calculator is programmed to generate a random integer between 1 and 10. In six trials, what is the probability that the number 3 is generated exactly twice?

Answer 1

this is binomial ... three (t) or not three (n)

p(t) = .1 ... p(n) = .9

(n + t)^6 = n^6 + 6 n^5 t + 15 n^4 t^2 + ... + 6 n t^5 + t^6

evaluate the 3rd term for the solution ... 15 * .9^4 * .1^2 = ?

Answer 2

There are only 8 numbers "between" 1 and 10
so the prob of getting a 3 = 1/8
prob of not getting a 3 = 7/8

prob(event as stated) = C(6,2) (1/8)^2 (7/8)^4
= 15(1/64)(2401/4096) = ...