Question 1030146
probability of winning is .55
probability of losing is 1 - .55 = .45
probability of winning in the third trial would be .45 * .45 * .55 = .111375 which rounds to .11.


your are using the binomial distribution.
that tells you how many successes you get out of 3 tries.
probabillity of success = .55
probability of failure = .45


that formula is p(x) = c(n,x) * p^x * q^(n-x)


what you entered should get you c(3,3) * .55^3 * .45^0 = .166375


what that is telling you is the probably of winning all 3 games out of 3.


that's not the same as winning only the third game.


i think your solution is .11.


it's the probability of losing the first 2 games times the probability of winning the third game.