Homework pointers: Probability

[BigPoppa]

Boiled down:

64 games played. Chance to win = 1/110 (0.00909...)

What I can't figure out is:

Chance to win EXACTLY 1 game. Not at least 1, exactly 1.

(1/110) x (1-1/110)^63 = ~0.005..., which may be correct, but

http://www.math.umt.edu/graham/math241/hw7assmt.pdf in #4D, he says the probability of winning exactly TWO games is 0.0946, but I can't figure out how to get that. Some pointers without the answer would be much appreciated.

P(exactly 2) = 1 - P(Win 0 + Win exactly 1 + win at least 3)?

If yes, how do I figure out P(win at least 3)?

Any pointers are appreciated.

LoKe, i'm not asking for answers, just help.

 

[DVK916]

I will tell you how he got 0.0946 and from that figure out how to do exactly one.

P(exactly 2) = (64 chose 2)*(1/110)^2*(109/110)^62=2016*(1/110)^2*(109/110)^62=0.0946.

 

[BigPoppa]

Thanks. I think i've got it now:

He gave us P(win exactly 2) because he hasn't taught us the nCr function yet. Should be soon.

P(At least 3) = 1 - P(Win 0 + win 1 + win 2).

Money.