Q. 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
I proceeded with the problem in the following way :
Sarah already remembers the first 2 digits of her ATM pin number. So we just have to restrict ourselves to the last 2 digits with the condition that they are greater than 5.
The probability that the 3rd digit is not greater than 5 = 5/9 (since it can be any number ranging from 1 - 9, so 9 options)
The probability that the 4th digit is not greater than 5 is also 5/9
Thus, probability that both 4th and 5th digits are not greater than 5 is 5/9*5/9 = 25/81
Therefore, probability that both the digits are greater than 5 in one turn will be 1 - 25/81 = 56/81
Hence, for 3 turns the probability will be 56/81 * 3 = 56/27
But the OA is C i.e 3/16
Can someone tell me where have I gone wrong?
Regards
MSD
(A) 1/2 (B) 1/4 (C) 3/16 (D) 3/18 (E) 1/32
I proceeded with the problem in the following way :
Sarah already remembers the first 2 digits of her ATM pin number. So we just have to restrict ourselves to the last 2 digits with the condition that they are greater than 5.
The probability that the 3rd digit is not greater than 5 = 5/9 (since it can be any number ranging from 1 - 9, so 9 options)
The probability that the 4th digit is not greater than 5 is also 5/9
Thus, probability that both 4th and 5th digits are not greater than 5 is 5/9*5/9 = 25/81
Therefore, probability that both the digits are greater than 5 in one turn will be 1 - 25/81 = 56/81
Hence, for 3 turns the probability will be 56/81 * 3 = 56/27
But the OA is C i.e 3/16
Can someone tell me where have I gone wrong?
Regards
MSD
When the going gets tough, the tough gets going.
