Question 1030146: Hi,
Can someone please help me with this.
The chances of James winning a game are 0.55. What is the probability of James winning the game in 3rd trial?
0.11
0.42
0.16
0.33
I have got the answer as 0.16 but not confident. I have used BINOM.DIST(3,3,0.55,FALSE) in excel.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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