The scenario of getting a red marble in the second draw could be like this:
1) no red in first draw and a red in the second draw
2) a red in the first draw and a red in the second draw

1) (7/9)*(2/8)=7/36
2) (2/9)*(1/8)=1/36

So, the probability that the second marble is drawn will be red = (7/36)+(1/36)=2/9

The answer is D.

If a red marble is selected first and a red marble is selected second, the probability is:

2/9 x 1/8 = 2/72

If a red marble is not selected first and a red marble is selected second, the probability is:

7/9 x 2/8 = 14/72

So the overall probability is 2/72 + 14/72 = 16/72 = 2/9.

Answer: D

This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something awful.
Someone would hold up n pieces of grass (for n people), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece would be minimized, since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n people had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected a straw.

The same applies to the original question here.

2 of the 9 marbles are red, so P(red ball selected SECOND) = P(red ball selected FIRST) = 2/9

Cheers,
Brent

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