Probability of shoot-outs (NHL)

[Kelsey72]

Can anyone help me with this problem.

In the last two years 12.56% (309/2460) of NHL games have gone to a shootout. If there are 9 games on one day, what is the probability that at least 3 go to a shootout?

Thanks,
K

[sixhigh]

<font color="#FF4444">[*] </font> B(9,3,0.1256)+..+B(9,9,0,1256) = 1.9%

This sums up the probability of 3, 4, .. and 9 games going to a shootout.

Where B(n,k,p) is the binomial distribution [ = C(n,k)*k^p*(n-k)^(1-p) ]

[AaronBrown]

sixhigh's method is correct, but he forgot to add exactly 3 shootouts, so his answer is for 4 or more instead of 3 or more. It's 9.3% for 3 or more shootouts (assuming they are independent).

[Kelsey72]

Thanks sixhigh and Aaron. Much appreciated.

K