# Thread: a basket contains 5 red balls, 3 blue balls, 1 green balls

1. ## a basket contains 5 red balls, 3 blue balls, 1 green balls

a basket contains 5 red balls, 3 blue balls, 1 green balls

what is the probability of getting 2 red balls, 1 blue ball and 1 green ball if a sample of 4 balls are taken from the basket.

How do i start solving this problem.

2. Originally Posted by ilikec
a basket contains 5 red balls, 3 blue balls, 1 green balls

what is the probability of getting 2 red balls, 1 blue ball and 1 green ball if a sample of 4 balls are taken from the basket.

How do i start solving this problem.

Hi "i like c",

there are 2 ways to work it out, if not more...

If a sample of 4 balls are taken,
we can list the order of selection if they are chosen one-by-one.

RRBG red 1st, red 2nd, blue 3rd, green 4th.... $\displaystyle P=\frac{5}{9}\ \frac{4}{8}\ \frac{3}{7}\ \frac{1}{6}$
RRGB
RBRG
RBGR
RGRB
RGBR
GRRB
GRBR
GBRR
BRRG
BRGR
BGRR

If you calculate the remaining probabilities and sum them, you will have the
probability of choosing 2 reds, 1 blue and 1 green.

Alternatively there are $\displaystyle \binom{9}{4}$ ways to choose 4 from 9.

There are $\displaystyle \binom{5}{2}$ ways to choose 2 reds

There are $\displaystyle \binom{3}{1}$ ways to choose one blue and only 1 way to choose the green.

Any 2 reds can go with any blue, hence we multiply those numbers of ways
to find the total number of ways we can get 2 reds, 1 blue and 1 green

Hence you divide that number by the total number of ways to choose 4 from 9.

3. How can you do it using Hyper-geometric Distribution principles?

4. Using the multivariate hypergeometric distribution makes the problem much easier in my opinion.

First let's label everything:

N= total number of balls = 9
n= total number of balls selected =4
R = total number of red balls = 5
B = total number of blue balls = 3
G = total number of green balls =1
r = total number of red balls selected = 2
b = total number of blue balls selected = 1
g = total number of green balls selected = 1

Now simply solve:

$\displaystyle \frac{{R\choose r}{B\choose b}{G\choose g}}{{N\choose n}}$

As you can see, it's simply the combination of ways to pick 2 red balls out of 5 red balls multiplied by the combination of ways to pick 1 blue ball out of 3 blue ball multiplied by the combination of ways to select 1 green ball from 1 green ball all divide by the combination of ways to select 4 balls from 9 balls.