There are 4*4 = 16 distinct ways of getting the last two digits. e.g. 68 or 97 or 79 etc. Only 1 of these 16 is correct.

On the first try, the probability of failing is 15/16 (say you punched in 68 and that was wrong)
Now you have 15 combinations remaining so logically, you will punch in one of the other 15. Probability of failing = 14/15. Say you punched in 97 and that was wrong.
Now you have sidelined 2 of the 16 cases and will punch in one of the remaining 14. Probability of failing = 13/14

Probability fo failing on all three tries = (15/16)*(14/15)*(13/14) = 13/16

Probability of succeeding = 1 - 13/16 = 3/16
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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.

Shouldn't it be 721/4096
using 1- (15/16)^3

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).
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