# Balls in a bag

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• Mar 3rd 2011, 11:09 AM
kandyfloss
Balls in a bag
There are 10 red balls, 10 green balls and 6 white balls.
Two balls picked at random - what is the probability that they are of different colors?

so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?
• Mar 3rd 2011, 11:20 AM
Quacky
The probability of each ball being picked is $\frac{1}{26}$, which is a start.

The way I would solve this would be to find the probability of two red balls, the probability of two green balls, and the probability of two white balls. If you add these, you will find the probability of receiving two balls that are the same colour.

I'm going to assume (as you haven't specified) that the balls are not replaced, and that this is therefore conditional probability.

P(2 red) = $\frac{10}{26}\times\frac{9}{25} = \frac{9}{65}$. This is because there are $10$ red balls out of the $26$ total during the first selection, and then there are $9$ red balls left out of the $25$ remaining balls during the second selection.
P(2 green) is the same as p(2 red) because there are the same amount of red balls as green balls, so p(2 green) $=\frac{9}{65}$
P(2 white) = $\frac{6}{26}\times\frac{5}{25}=\frac{3}{65}$

Do you understand so far?

So what is the total probability of receiving two balls that are the same colour?

And, therefore, how can you work out from that the probability of receiving two balls which are not the same colour? Show your working if you get stuck.
• Mar 3rd 2011, 11:36 AM
mr fantastic
Quote:

Originally Posted by kandyfloss
There are 10 red balls, 10 green balls and 6 white balls.
Two balls picked at random - what is the probability that they are of different colors?

so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?

Draw a tree diagram.
• Mar 3rd 2011, 11:55 AM
kandyfloss
So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
Thank you.
• Mar 3rd 2011, 12:02 PM
Plato
Quote:

Originally Posted by kandyfloss
There are 10 red balls, 10 green balls and 6 white balls. Two balls picked at random - what is the probability that they are of different colors?

Probability both are white is $P(W_1\cap W_2)=\frac{6}{26}\cdot\frac{5}{25}$.

So find $P(R_1\cap R_2)+P(G_1\cap G_2)+P(W_1\cap W_2)$
• Mar 3rd 2011, 12:29 PM
Archie Meade
Quote:

Originally Posted by kandyfloss
So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
Thank you.

You could have tried...

What is the probability of getting

(1) a red with a green

or

(2) a red with a white

or

(3) a green with a white

There are $\binom{26}{2}$ ways to choose 2 balls. This is the denominator for your probability fraction.

There are 10(10)=100 ways to get a green with a red

There are 10(6)=60 ways to get a red with a white

There are 10(6)=60 ways to get a green with a white

The sum of the above three products is the numerator of your probability fraction.
• Mar 3rd 2011, 01:13 PM
kandyfloss
I went through, both of your methods. Since I already did it using the first one, i'll stick to it.
But thanks you very much :) it really helped me understand better.
• Mar 3rd 2011, 06:42 PM
Wilmer
[1]Prob white : 6/26 = 3/13 ; [2]prob not white : 20/25 = 4/5 ; 3/13 * 4/5 = 12/65
[1]Prob not white: 20/26 = 10/13 ; [2]prob not same: 16/25 ; 10/13 * 16/25 = 32/65

12/65 + 32/65 = 44/65