First two digits are already known. For each of the last two digits, we have 4 options: 6, 7, 8 or 9.
So the last two digits can take a total of 4 * 4 = 16 values.

For a total of 16 possibilities, Sarah has three tries. So three things could happen:

1. She guesses on the first try, for which the probability is 1/16

2. She doesn't guess on first but second try. So in first try she entered a wrong code and in second try she entered a correct code.
probability of the same = 15/16 * 1/15 = 1/16.

3. First two tries wrong code, third try correct code. Probability = 15/16 * 14/15 * 1/14 = 1/16.

So total probability = 1/16 + 1/16 + 1/16 = 3/16.

Hence answer is C.

chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis
Please explain why Sahilkm 's reasoning is wrong

Sarah cannot completely remember her four-digit ATM pin number. She does remember the first two digits, and she knows that each of the last two digits is greater than 5. The ATM will allow her three tries before it blocks further access. If she randomly guesses the last two digits, what is the probability that she will get access to her account?

(A) 1/2
(B) 1/4
(C) 3/16
(D) 3/18
(E) 1/32

Since Sarah has three trials left and only one combination of the last two digits is going to be correct. It is possible to get correct one punched on the very first attempt or on the second, if not then on third.
So, what we multiple it by 3(for three trials).

Last two digits(both > 5) can be one of 6,7,8 & 9.
Third digit can have 4 distinct digit and similarly fourth digit can be one of the four digits(code can have same digits for both third and fourth position).
Thus, total possibilities of combinations = 4x4 = 16

Probability = \(\frac{1}{16}\)*3

Hence Answer is (C).
_________________