Question

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

Answer 1

To find the probability you will multiply the probability of each possibility.

2/12 (1st attempt) x 1/11 (2nd attempt)

The answer is 1/66 chance (Choice C).

Answer 2

Answer: C 1/66

Step-by-step explanation:

P(letter O both times)=P(letter O on first trial)×P(letter O on second trial)

P(letter O on first trial)=total number of O/total number of letters

=2/12

=1/6

now he doesn't replace it

so, P(letter O on second trial)= total number of O left/total number of alphabets left

= 1/11

so,P(letter O on both trials)=1/66