A bag contains tiles with the letters -C-O-M-B-I-N-A-T-I-O-N-S. Lee chooses a tile without looking and doesn't replace it. He chooses a second tile without looking. What is the probability that he will choose the letter O both times?

A) 1/132
B) 1/72
C) 1/66
D) 1/23

Is the answer B?

7 answers

prob = (2/12)(1/11) = 1/66

answer is 1/66 thanks reiny

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To find the probability that Lee will choose the letter O both times, we need to consider two factors: the total number of tiles in the bag and the number of tiles with the letter O.

First, let's determine the total number of tiles in the bag. We are given the letters: C, O, M, B, I, N, A, T, I, O, N, S.

Counting the letters, we have a total of 12 tiles.

Next, let's determine the number of tiles with the letter O. We see that there are 2 O's in the given letters.

Now, since Lee chooses a tile without replacing it, the number of available tiles decreases with each pick. After Lee chooses the first tile, there will be a total of 11 tiles left in the bag, and only 1 O left.

So, the probability of Lee choosing an O on his first pick is 2/12 (which simplifies to 1/6), and the probability of Lee choosing an O on his second pick is 1/11.

To find the probability of both events occurring (Lee choosing an O on both picks), we multiply the probabilities together:

(1/6) * (1/11) = 1/66.

Therefore, the correct answer is C) 1/66.