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?
A. one over fifty-five
B. two over fifty-five
C. two over one hundred twenty-one
D. three over one hundred twenty-one.
There are 11 tiles in the bag. The probability of choosing an I on the first draw is 2/11, since there are two I's in the bag. After one I has been chosen and not replaced, there is only one I left out of 10 tiles, so the probability of choosing an I on the second draw is 1/10. Therefore, the probability of choosing I both times is:
(2/11) * (1/10) = 2/110 = 1/55
So the answer is (A) one over fifty-five.
To find the probability that Tyler will choose the letter "I" both times, we can use the following steps:
Step 1: Determine the total number of tiles in the bag. In this case, there are 11 tiles since there are 11 letters in the word "PROBABILITY".
Step 2: Determine the number of tiles with the letter "I". In this case, there are 3 tiles with the letter "I".
Step 3: Since Tyler doesn't replace the tile after choosing it, the total number of tiles decreases by 1 each time. Therefore, the probability of choosing "I" for the first time is 3/11 (3 "I" tiles out of 11 total tiles).
Step 4: After Tyler chooses the first tile, there are now 10 tiles left in the bag, with only 2 of them being "I" tiles. Thus, the probability of choosing "I" for the second time is 2/10 (2 "I" tiles out of 10 total tiles).
Step 5: To find the probability of both events occurring, we multiply the probabilities of each individual event together:
(3/11) * (2/10) = 6/110 = 3/55
Therefore, the probability that Tyler will choose the letter "I" both times is three over fifty-five (Option D).