an urn contains 8 orange balls and 7 green balls. If 7 balls at random are chosen what is probability that he will select 3 orange balls and 4 green

Answer 1

Since we begin with an urn that has 15 balls (8 orange balls and 7 black balls) and then select 7 balls at random from this urn, we'll need to use the hypergeometric distribution to properly calculate the probability of selecting 3 orange balls and 4 green balls (urn models like this current problem are classic examples of hypergeometric distributions) from our original urn. If we let X denote the number of orange balls in a sample of size 7 drawn from our original urn, then X has a hypergeometric distribution given by

P(X=x|15,8,7)=\frac{{8\choose x}{7\choose 7 - x}}{{15\choose 7}},

x = 0,1,\ldots,7 and 8-(15-7)\leq x\leq 8 (i.e. 0\leq x \leq 8)

We can now calculate that the probability that X = 3 (i.e. the probability of randomly selecting 3 orange balls and 4 green balls from our original urn) is

P(X=3|15,8,7)=\frac{{8\choose 3}{7\choose 7 - 3}}{{15\choose 7}}

=\frac{{8\choose 3}{7\choose 4}}{{15\choose 7}}

=\frac{56*35}{6435}

=\frac{1960}{6435}

=\frac{392}{1287}

\approx 0.3046