probability question

Question

A bag contains five red balls and four green balls. Another bag contains three red balls and seven green balls. A ball is drawn from the first bag, and three balls are drawn from the other bag. Find the probability that out of four balls drawn, three are red and one is green.

Answer 1

This is a hypergoemetric distribution problem.  The pdf for a hypergeom is

(A nCr x)(B nCr n-x)/(A + B nCr n),    x=0,1,...,n

where A and B are the number of objects in the bag of each type and x is the number of objects of type A selected with a sample size n.

We see two cases either a red ball may be pulled from the first bag or a green ball may be pulled from the first bag.  Let

A_i = red balls

B_i = green balls

x_i = number of red balls chosen

n_i = sample size

where the subscript i tells us which bag (i=1 or i=2).

Let A_1 = 5, b_1 = 4, n_1 = 1, A_2 = 3, B_2 = 7, n_2 = 3.  In the first case x_1 = 1 and x_2 = 2 and in the second case x_1 = 0 and x_2 = 3.

 

Since bag 1 and 2 are independent of each other we look at the product of the probabilities of the two bags. 

In case 1,

∏ (A_i nCr x_i)(B_i nCr n_i - x_i)/(A_i + B_i nCr n_i)

=(5 nCr 1)(r nCr 1 - 1)/(5 + 4 nCr 1) * (3 nCr 2)(7 nCr 3 - 2)/(3 + 7 nCr 3)

=5/9 * 21/120 = 7/72

 

In case 2,

(5 nCr 0)(4 nCr 1 - 0)/(5 + 4 nCr 1) * (3 nCr 3)(7 nCr 3 - 3)/(3 + 7 nCr 3)

=4/9 * 1/120 = 1/270.

 

Since the two cases are disjoint events, we can add them together to get,

7/72 + 1/270 = 109/1080.