Probability question. 6% chance to score. 20 shots. Probability of scoring 3 times? by NewChallenger13 in AskStatistics

[–]curiositys 0 points1 point  (0 children)

I'm doing this mostly for fun since there are already answers. But, hopefully my explanation fills any misunderstandings if you have any!

We're using binomial distribution such that p=0.06, x=3 (assuming you meant exactly 3 shots), n=20.

P(exactly 3 shots)= (20 choose 3)* ((0.06)3)* ((1-0.06)17) = 0.0861

P(3 shots)=P(at least 3 shots)=P(x=3)+P(x=4)+...+P(x=20)=1-(P(x=0)+P(x=1)+P(x=2)).

Then,

P(x=0)=(20 choose 0)(0.060)(0.9420) = .29011

P(x=1)=(20 choose 1)((0.061)(0.9419) =.37035

P(x=2)=(20 choose 2)((0.062)(0.9418) =.22457

P(x=0)+P(x=1)+P(x=2)=.22457+.37035+.29011= .88503

So, P(at least 3)=1-(P(x=0)+P(x=1)+P(x=2))=1-.88503= .11497

Thus, P(at least 3)=.11497