PLEASE HELP!
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 9 and 8. What is the probability that the third digit is 7?
A. 1/9
B. 1/3
C. 1/8
D. 1/7
I think B or D but if i choose i would go with D
You have the choice of 9 digits, since 0 is not used.
Two of those are already gone, so .....
(I must be 100% sure of your answer)
To solve this problem, we need to find the total number of possible outcomes and the number of favorable outcomes.
The first two digits are fixed: 9 and 8. For the third digit to be 7, we need to choose one digit from the remaining 7 non-zero digits (excluding 9 and 8) from 1 to 7.
The total number of possible outcomes is the number of ways to choose one digit from 1 to 7, which is 7.
The number of favorable outcomes is 1 (since we want the digit to be 7).
Therefore, the probability is favorable outcomes divided by total outcomes:
P(Third digit is 7) = 1/7.
So, the correct answer is option D: 1/7.