Sarah cannot completely remember her four-digit ATM pin number

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).

Solution:

The choices Sarah has for the last 2 digits are 6, 7, 8 and 9

So, the total number of possible last 2 digits are \(= 4\times 4 = 16\)

Probability that she gets the correct last 2 digits \(= \frac{1}{16 }\)

Probability that she gets the correct last 2 digits in the 3 tries

= (Probability that she gets it in 1st try) OR (Probability that she gets it in 2nd try) OR (Probability that she gets it in 3rd try)

= (Probability that she gets it in 1st try) + (Probability that she gets it in 2nd try) + (Probability that she gets it in 3rd try)

\(= \frac{1}{16} + \frac{1}{16} + \frac{1}{16}\)

\(= \frac{3}{16}\)

Hence the right answer is Option 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

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