Probability Question

[karim.nazmy]

Hi,

I need help solving the below question using the combination approach.

A box contains 4 green balls, 3 red balls, and 2 yellow balls. If 2 balls are selected without replacement, what is the probability that they will be both the same color?

A) 2/9

B) 20/81

C) 5/18

D) 1/3

E) 5/12


My approach is as follows:
There is a total of 9C2 combinations which are 36. To get 2 of the same color, there are 2 green combinations, 1 red and 1 yellow. Therefore, answer would be (2+1+1)/36 = 4/36= 2/9 which is incorrect as the answer here is C.

[Bullzi]

Hello,

This is how I approached the problem,

Total possible combinations - 9C2 = 36
Choosing 2 Green balls can be done in 4C2 ways = 6
Choosing 2 Red balls can be done in 3C2 ways = 3
Choosing 2 Yellow balls can be done in 2C2 ways = 1

Considering all values - [spoiler]probability = (6+3+1)/36 = 5/18 Option C[/spoiler]

Thanks,
Bullzi

[talaangoshtari]

Hi karim.nazmy,

Here's another approach:

We can consider 3 scenarios:

g:green
y:yellow
r:red

gg + yy + rr

(4/9)(3/8) + (2/9)(1/8) + (3/9)(2/8) = 5/18

[ceilidh.erickson]

My approach is as follows:
There is a total of 9C2 combinations which are 36. To get 2 of the same color, there are 2 green combinations, 1 red and 1 yellow. Therefore, answer would be (2+1+1)/36 = 4/36= 2/9 which is incorrect as the answer here is C.

Your math doesn't quite work here. There aren't simply 2 green combinations when you have 4 balls. You need to consider every combination: green1+green2, green1+green3, green1+green4, etc.

In fact, for 4C2, there are 4!/((2!)(2!)) = 6 ways of choosing 2 of the 4 green balls. Then there are 3 ways to choose 2 out of 3 red balls: 3!/((2!)(1!)). Then there is, as you said, only 1 way to choose both yellow balls. So, there are 6 + 3 + 1 = 10 total ways of choosing 2 of the same color out of 36 total ways of choosing 2 balls of any color. 10/36 = 5/18.

The answer is C.