Problem Solving

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

there is only one correct pin, you are assuming if both the last 2 digits are greater than 5, they will be correct. but last 2 digits can well be greater than 5 and still not correct
the last 2 digits can be same also, each digit has 4 options (6,7,8,9) so they can be guessed in (4*4)=16 ways. Of these only one sequence is correct. The probability to get it correct is 1/16 for any one turn
for 3 chances, the probability will be 3/16

scoobydooby wrote:there is only one correct pin, you are assuming if both the last 2 digits are greater than 5, they will be correct. but last 2 digits can well be greater than 5 and still not correct
the last 2 digits can be same also, each digit has 4 options (6,7,8,9) so they can be guessed in (4*4)=16 ways. Of these only one sequence is correct. The probability to get it correct is 1/16 for any one turn
for 3 chances, the probability will be 3/16

Your explaination for first part, that probability of selecting correct pin is correct

P(correct pin) = 1/16, hence P(incorrect pin) = 15/16

However, Now to get access, the user has maximum 3 chances, so the events are

Either

(1) he enters correct pin in ATM first time

or

(2) he enters incorrect pin first time and correct pin second time

or

(3) he enters incorrect pin first and second time and correct pin third time

so,

P(get access) = P(Event1) + P(Event2) +P(Event3)
=(1/16) + (15/16 * 1/16) + (15/16 * 15/16 * 1/16)

nitin86, am getting 45/16^2 after simplying the last step in your solution, doesnt tie with any choices. do agree with your approach though

Lets use the (1-p) rule. There are 4*4 ways of picking the last two numbers( 6,7,8,9) and there are three ways the pin will be wrong IE

WW
CW
WC

3/16

Austin 3:16 does any wrestling fan remember that...

jayjk78 wrote:Lets use the (1-p) rule. There are 4*4 ways of picking the last two numbers( 6,7,8,9) and there are three ways the pin will be wrong IE

WW
CW
WC

3/16

Austin 3:16 does any wrestling fan remember that...

jayjk78, but arent we supposed to find the prob of getting the pin right? in which case it would be 13/16?