So I have this question:
a box contains 8 balls
2 red and 6 black, 2 balls are drawn in succession and with replacement
the answer is:
(2/8)(2/8) + (6/8)(6/8)
now, i have another question:
3 balls are drawn in succession we have 3 white 4 red and 5 green balls, we win 1 dollar when we get a white ball and lose 1 dollar when we get a red ball and 0 if we get a green ball.
what's the probability of getting 0 dollars?
I did this thinking the problem was of the same type:
we need 3 green or 1 white, 1 red and 1 green without replacement
(5/12)(4/11)(3/10) + (3/12)(4/11)(5/10)
however, this is false.
notice how we could word the question almost the same way.
Best Answer
You didn't say what the first question was, but it appears to be the probability the two balls match. In that case you are correct as you have accounted for all the possibilities. For the second, you have calculated the chance of three green (not white as you say in the text), and white, red, green, but there are five more orders for the three colors. Your second term should be multiplied by $6$.