## Statistics, ping pong balls in bag probability question

1. The problem statement, all variables and given/known data

Twelve colored ping-pong balls are placed into a shopping bag and well mixed. There are two red balls, six blue balls and four green balls. One ball is selected at random, its color noted and then it is set aside. A second ball is then randomly selected and its color noted. What is the probability that both balls are the same color?

2. Relevant equations

P(two same color) = P(red) + P(blue) + P(green)

3. The attempt at a solution

P(two same color) = (2/12 + 1/12) * (6/12 + 5/12) * (4/12 + 3/12) = .8125
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Hi Calculator14!
 Quote by Calculator14 P(two same color) = P(red) + P(blue) + P(green) 3. The attempt at a solution P(two same color) = (2/12 + 1/12) * (6/12 + 5/12) * (4/12 + 3/12) = .8125
I don't understand

and eg what is 6/12+5/12 supposed to be?
Start again …

what is P(both balls are blue)?
 Hi tiny-tim! I'm sorry, I am lost on how to start this equation. Would you be able to guide me a bit through this? Thank you for your help!!!

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## Statistics, ping pong balls in bag probability question

Hi Calculator14!

(just got up …)

If you select two balls out of 12, what is the probability that both balls are blue?

 Tags probability, statistics