Probability question : Need expert help

[GMATMadeEasy]

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?

[maihuna]

To me it is coming out to be 3/16 assuming she tries differently in all three times:

There are a total of 4*4=16 possibility.

Failure in 1st try = 15/16, in 2nd, 14/15, in 3rd 13/14

Total chance of failure = (15/16)*(14/15)*(13/14) = 13/16

So chances of success : = 1-13/16 = 3/16

I may be wrong though in case I am missing some key principle of independent events.

[Night reader]

source of this query is Nova's
@maihuna, the logic of this question and its OA (3/16) are counter-verse with my logic and solution. I got answer 1/4, and I hope this approach will withstand critiques here...

pin=ABCD, P(A)=P(B)=100% OR 1
total possibilities for CD (order matters)=4P2= 4!/2!=24/2=12
1st trial, P(success) = 1*1*(1/12) = 1/12 Yes, if No 11/12 proceed to below
2nd trial, P(success) = 1*1*(1/11) = 1/11 Yes, if No 10/11 proceed to below
3rd trial, P(success) = 1*1*(1/10) = 1/10

total probabilities =1/12 + (11/12)*(1/11) + (11/12)*(10/11)* 1/10 = 3/12 OR 1/4

I am considering the way Sarah guesses on her pin-code is an educated guess that is she first arranges the set of digits out of possible 4 {6,7,8,9} and she should get the total of 12 possible arrangements. Then every time she has success or failure in the selection of pin-code she either gets access to money (account) or crosses out the unsuccessful set (of two digits). However the problem says that Sarah randomly guesses her last two digits - she's nervous she wants money badly she can't handle permutations - enough she does random way or simple 4*4 = total 16 ways every time.

To me it is coming out to be 3/16 assuming she tries differently in all three times:

There are a total of 4*4=16 possibility.
Success by Sarah: 1st, 2nd, 3rd trials --> (1/4)*(1/4) OR 1/16; Failure by Sarah: 1st, 2nd, 3rd trials --> 15/16
simple calc (1/16)*3=3/16
Failure in 1st try = 15/16, in 2nd, 14/15, in 3rd 13/14 <-- victim of Nova's Sarah logic

Total chance of failure = (15/16)*(14/15)*(13/14) = 13/16

So chances of success : = 1-13/16 = 3/16

I may be wrong though in case I am missing some key principle of independent events.

[garuhape]

This should provide the solution:



Uploaded with ImageShack.us[/img]

I hope it's correct though!

[Night reader]

in the first trial from the nice drawing provided by you we read 1/16 i.e. 1 selection out of 4 possible (1/4)*(1/4). What are 1/15 and 1/14 in the second and the third trials?

This should provide the solution:



Uploaded with ImageShack.us[/img]

I hope it's correct though!

[garuhape]

in the first trial from the nice drawing provided by you we read 1/16 i.e. 1 selection out of 4 possible (1/4)*(1/4). What are 1/15 and 1/14 in the second and the third trials?

This should provide the solution:



Uploaded with ImageShack.us[/img]

I hope it's correct though!

On the first attempt, there are 15 in sixteen times to miss-guess. Later, on the second attempt, there are only 14 wrong attempts out of 15 left given the case, that the first attempt was wrong. And so on.

I'm not sure whether it's write though.

[kevincanspain]

source of this query is Nova's
@maihuna, the logic of this question and its OA (3/16) are counter-verse with my logic and solution. I got answer 1/4, and I hope this approach will withstand critiques here...

pin=ABCD, P(A)=P(B)=100% OR 1
total possibilities for CD (order matters)=4P2= 4!/2!=24/2=12
1st trial, P(success) = 1*1*(1/12) = 1/12 Yes, if No 11/12 proceed to below
2nd trial, P(success) = 1*1*(1/11) = 1/11 Yes, if No 10/11 proceed to below
3rd trial, P(success) = 1*1*(1/10) = 1/10

total probabilities =1/12 + (11/12)*(1/11) + (11/12)*(10/11)* 1/10 = 3/12 OR 1/4

I am considering the way Sarah guesses on her pin-code is an educated guess that is she first arranges the set of digits out of possible 4 {6,7,8,9} and she should get the total of 12 possible arrangements. Then every time she has success or failure in the selection of pin-code she either gets access to money (account) or crosses out the unsuccessful set (of two digits). However the problem says that Sarah randomly guesses her last two digits - she's nervous she wants money badly she can't handle permutations - enough she does random way or simple 4*4 = total 16 ways every time.

To me it is coming out to be 3/16 assuming she tries differently in all three times:

There are a total of 4*4=16 possibility.
Success by Sarah: 1st, 2nd, 3rd trials --> (1/4)*(1/4) OR 1/16; Failure by Sarah: 1st, 2nd, 3rd trials --> 15/16
simple calc (1/16)*3=3/16
Failure in 1st try = 15/16, in 2nd, 14/15, in 3rd 13/14 <-- victim of Nova's Sarah logic

Total chance of failure = (15/16)*(14/15)*(13/14) = 13/16

So chances of success : = 1-13/16 = 3/16

I may be wrong though in case I am missing some key principle of independent events.

Are you assuming that the last two digits of her PIN must be distinct?

[Night reader]

Hi Kevin, I'll pass
wrong assumption was made 4*3 <-- distinct numbers
total possibilities=4*4 OR 16
thanks

[GMATMadeEasy]

Could some one solve the question please.

[aleph777]

There's a lot of filler in that question to make it sound more complicated, but all we're really being asked is what is the probability that she picks two numbers correctly.

We know that each number is greater than 5, meaning they could only be 6 7 8 or 8. So there is a 1 in 4 chance she'll pick each number correctly. Because she needs to pick BOTH correctly, though, we multiply the probabilities first.

1/4 * 1/4 = 1/16

And then, since she has three chances to get the number right, we multiply 3 * 1/16. Which gives us 3/16.

[tomada]

In this case, we're assuming that she might enter the same, incorrect two digits on three consecutive tries.
Couldn't we assume that, if her first try isn't correct, she won't repeat the same two digits on her second try and, if her two digits on her second try are incorrect, that she won't select either of her first two combinations for the third and final try?

Thus, I think the probability should be 1/16 + 1/15 + 1/14 ... (I'm too lazy to find the LCD)

There's a lot of filler in that question to make it sound more complicated, but all we're really being asked is what is the probability that she picks two numbers correctly.

We know that each number is greater than 5, meaning they could only be 6 7 8 or 8. So there is a 1 in 4 chance she'll pick each number correctly. Because she needs to pick BOTH correctly, though, we multiply the probabilities first.

1/4 * 1/4 = 1/16

And then, since she has three chances to get the number right, we multiply 3 * 1/16. Which gives us 3/16.

[Night reader]

lol, what about P(success-1) + P(failure-1)byP(success-2) + P(failure-1)byP(failure-2)byP(success-3)
1/16 + (15/16)*1/15 + (15/16)(14/15)*1/14
Tomada, we get 3/16 eventually

In this case, we're assuming that she might enter the same, incorrect two digits on three consecutive tries.
Couldn't we assume that, if her first try isn't correct, she won't repeat the same two digits on her second try and, if her two digits on her second try are incorrect, that she won't select either of her first two combinations for the third and final try?

Thus, I think the probability should be 1/16 + 1/15 + 1/14 ... (I'm too lazy to find the LCD)

There's a lot of filler in that question to make it sound more complicated, but all we're really being asked is what is the probability that she picks two numbers correctly.

We know that each number is greater than 5, meaning they could only be 6 7 8 or 8. So there is a 1 in 4 chance she'll pick each number correctly. Because she needs to pick BOTH correctly, though, we multiply the probabilities first.

1/4 * 1/4 = 1/16

And then, since she has three chances to get the number right, we multiply 3 * 1/16. Which gives us 3/16.