Probabilty

[CSASHISHPANDAY]

n a certain game, a player begins with a bag containing tiles numbered 1 through 10, each of which has an equal probability of being selected. The player draws one tile. If the tile is even, the player stops. If not, the player draws another tile without replacing the first. If this second tile is even, the player stops. If not, the player draws a third tile-without replacing either of the first two tiles-and then stops. What is the probability that at the conclusion of the game, the sum of the tiles that the player has drawn is odd?

(A)5/18


(B) 13/36


(C)3/8


(D)23/36


(E)5/8

[NicoleWhite]

Hi. I think the answer is B.

I believe there are two ways to end up with a sum that is odd: draw an odd first, then an even, and stop, or draw all three odds. Any other way will end up with an even sum.

The probability of the first is: P(odd) * P(even) = 5/10 * 5/9 = 1/2 * 5/9 = 5/18
The probability of the second is: P(odd) * P(odd) * P(odd) = 5/10 * 4/9 * 3/8 = 12/144 = 1/12

Adding these two together: 5/18 + 1/12 = 78/216 = [spoiler]13/36[/spoiler]

I am new here so please let me know if I've done something incorrect.