A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler 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 I both times?.
The probability of choosing the letter I on the first draw is $\frac{4}{12}$ (since there are 4 I's out of 12 letters total). Since Tyler doesn't replace the first tile, there are now 11 tiles left, but still 4 I's. Therefore, the probability of choosing I on the second draw is $\frac{4}{11}$. To find the probability of both events happening (choosing I twice), we multiply the probabilities: $$\frac{4}{12}\cdot \frac{4}{11}=\frac{16}{132}=\boxed{\frac{4}{33}}.$$
