Statistics conditional probability help??!?

Since we never know the colour of the first ball it is as if it was never drawn, it may as well still be in the bag hence:

prob = 2/6 = 1/3 <----

Any dependency the second ball has on the first is not important, since, after drawing the first we somehow still do not know what colour it is. This is a bit of a philosophical point because it is tricky to draw a ball and still not know what colour it is, we could "palm" the ball: covering it up would allow us to draw without knowing the colour but then we may argue that drawing without knowing the colour doesn't constitute a draw at all. If we leave the ball in the bag then draw a second ball we can claim that we have not drawn the first ball since the first ball *must* be left in the bag, and because we don't know the colour of the first ball we can claim that it is either of the two colours remaining in the bag.

If this argument is a bit too much to contemplate then we could calculate by adding probabilities of p(white then red) = (4/6)*(2/5) = 8/30 and p(red then red) = (2/6)*(1/5) = 2/30

8/30 + 2/30 = 10/30 = 1/3 <----

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Note that if you had only 1 red ball and 4 white then the first argument wouldn't hold: If the second ball is a red then in this example the first ball *must* be a white. We could not "leave the first ball in the bag" then claim later that it is any of the two colours not picked as the first colour had to be white. In this case the probability is (4/6)*(1/5) = 4/30 = 2/15.

P(second ball is red) = P(first ball red, second ball red) + P( first ball not red, second ball red)

= (2/6)(1/5) + (4/6)(2/5)

= 2/30 + 8/30 = 10/30

= 1/3