Question

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?

Answer 1

-- "three non-zero digits";
so the last one can't be zero

-- "digits cannot be repeated" and "the first two digits are 9 and 8";
so the last one can't be 9 or 8

-- What's left ?
1, 2, 3, 4, 5, 6, or 7 . . . . . . . Seven possibilities.

-- Probability = (number of possible successes) / (number of possibilities)

That's 1/7 . (choice-D)

Answer 2

The calculated probability that the third digit is 7 is 1/8

What is the probability that the third digit is 7?

From the question, we have the following parameters that can be used in our computation:

• Digits cannot be repeated
• First two are 9 and 8

Since the first two digits are 9 and 8, then we have

Digits left = 8

Digit 7 = 1

So, the probability that the third digit is 7 is

P = 1/8

Hence, the probability that the third digit is 7 is 1/8