Thread: 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.

Please help.

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.

Please help.

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....
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 ways to choose 4 from 9.

There are ways to choose 2 reds

There are 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.

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

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:



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.