Problem Solving

Hi All,

Can someone please help me understand how to solve the below question:

==============================================
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

OA = C

Thanks
Mohit

goelmohit2002 wrote:Hi All,

Can someone please help me understand how to solve the below question:

==============================================
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


Thanks
Mohit

For probability questions, it's always a good idea to start by writing down the general formula:

Probability = (# of desired outcomes)/(total # of possibilities)

Let's start with # of desired outcomes. Sarah gets 3 guesses, so she has 3 chances to match.

Next, let's calculate total # of possbilities. Each of the last two digits is greater than 5, so must be 6, 7, 8 or 9. Since there are 4 possibilities for each digit, there are 4*4 = 16 total possible pairs of digits.

So, her probability of gaining access is 3/16... choose C.

She knows the first 2 digits and she knows the last two digits are over 5.

So the last two digits will either bet 6, 7, 8, or 9.

So

4 * 4 = 16 combinations

3 chances to get the combinations...

so

C. 3/16 is the probabality.

Hi Stuart,

Thanks.

But isn't 2nd guess depenedent on the outcome of the first....basically what I mean is this:

If Sarah will not get the access in the first attempt, then he will have only 15 numbers to choose from ( let's assume that she tried no. 12 first time, which was wrong)....so will her probabilty not dependent on the outcome of first attempt...so shouldn't her probaility for second guess = (15/16) * (1/15)

i.e. total probability = 1/16 + ((15/16)*(1/15)) + ((15/16)*(14/15)*(1/14)

Basically my doubt is that why are we assuming the outcomes independent of each other....i.e. 1/16 each time....

and total probability = 3 * (1/16) = 3/16.

instead of the above method.....

Thanks
Mohit

goelmohit2002 wrote:Hi Stuart,

Thanks.

But isn't 2nd guess depenedent on the outcome of the first....basically what I mean is this:

If Sarah will not get the access in the first attempt, then he will have only 15 numbers to choose from ( let's assume that she tried no. 12 first time, which was wrong)....so will her probabilty not dependent on the outcome of first attempt...so shouldn't her probaility for second guess = (15/16) * (1/15)

i.e. total probability = 1/16 + ((15/16)*(1/15)) + ((15/16)*(14/15)*(1/14)

Basically my doubt is that why are we assuming the outcomes independent of each other....i.e. 1/16 each time....

and total probability = 3 * (1/16) = 3/16.

instead of the above method.....

Thanks
Mohit

Hi Mohit

The last 2 diigts need not be the same.

She should not be able to discount any digit after a try.

Suppose the last 2 digits she chose in the first attempt were 6 and 7 in the same order and the pin is wrong .. Can she discard it ? No. What if the digits were in reverse order i.2 the pin was 1276 instead of 1267

or what if 6 or 7 occured with 8 or 9 in another order.

She will never be able to discard a digit and hence will have the same numbers of digits in the pool for any attempt.

To discard a digit she must try the permuation of the digit with any other possible digit.. We have only 3 attempts here. Hence not possible.

Stuart will be able to explain better.

Ian Stewart wrote:You're right, but you'll notice that, if you cancel, your answer above simplifies to 1/16 + 1/16 + 1/16. If she makes it to the second attempt, she's slightly more likely to guess her PIN, because she's ruled out one wrong number, but she's commensurately less likely to get to that second attempt in the first place, because she might guess right the first time.

A simpler version of the same problem: If you put two slips of paper in a hat, one with the letter A on it, and one with the letter B on it, and you pick the two slips in order, what's the probability the second slip is an 'A'? Of course it's 1/2. The same thing is true in the question above. Suppose she guesses all 16 different possible PIN numbers in some sequence. What's the probability the third guess is the right one? It's 1/16, of course - the same as for her first guess, or for her sixteenth guess.

Thanks a lot Ian.

Yes, the no. comes to be 1/16 in all first, second and third guess.....

But can you please tell...can this be generalised to a bigger no. cases...i.e. can we say that in the following type of question that the probability is 1/4 using the same logic as above, without calculating the same.....i.e. calculating till 10th outcome ?

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart ?"

a) 1/4
b) 1/5
c) 5/26
d) 12/42
e) 13/42

OA = A

goelmohit2002 wrote: Thanks a lot Ian.

Yes, the no. comes to be 1/16 in all first, second and third guess.....

But can you please tell...can this be generalised to a bigger no. cases...i.e. can we say that in the following type of question that the probability is 1/4 using the same logic as above, without calculating the same.....i.e. calculating till 10th outcome ?

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart ?"

a) 1/4
b) 1/5
c) 5/26
d) 12/42
e) 13/42

OA = A

This question is different for two reasons:

First, you've specified that the 10th card must be a heart. In the original question, she just had to get the password right on one of the three tries. To be analagous to your question, it would have been "what's the probability that she gets the password correct on the third try".

Second, there's more than 1 heart in the deck, which makes the question infinitely more complicated.

Here's a question that would be similar to the one in this thread:

Laura shuffles a regular deck of playing cards and then flips over the top 10 cards. What's the probability that one of the 10 cards is the 4 of spades?

The answer to that question would simply be 10/52.

Here's the general rule you can apply: if the object for which you're searching is unique and there's no replacement, you can treat the question as though you're making all of your attempts simultaneously.

However, the answer to the question you posted is also simply 1/4. We can show that to be the case using some basic logic.

There are an equal number of hearts, spades, clubs and diamonds in the deck. For there to be less than a 1/4 chance of the 10th card being a heart, there would have to be greater than a 3/4 chance of the 10th card being a club, spade or diamond (and greater than a 1/4 chance of the 10th card being some specific suit). Since from a probability point of view the suits are identical, that just makes no sense at all.