Statistics, ping pong balls in bag probability question

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Calculator14
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Homework Statement



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?

Homework Equations



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

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! :smile:
Calculator14 said:
P(two same color) = P(red) + P(blue) + P(green)

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 :redface:

why are you multiplying instead of adding?

and eg what is 6/12+5/12 supposed to be? :confused:

Start again …

what is P(both balls are blue)? :smile:
 
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!