N=10
n=3
x=2
k=4
Hypergeometric distribution
P(X)=(kx)(N−kn−x)(Nn)P(2)=(42)(10−43−2)(103)P(2)=6×6120=0.3P(2)=0.3P(X) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}} \\ P(2) = \frac{\binom{4}{2} \binom{10-4}{3-2}}{\binom{10}{3}} \\ P(2) = \frac{6 \times 6}{120} = 0.3 \\ P(2) = 0.3P(X)=(nN)(xk)(n−xN−k)P(2)=(310)(24)(3−210−4)P(2)=1206×6=0.3P(2)=0.3
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment