Lola forgot last three digits

[sanju09]

Lola forgot last three digits of one telephone number. All she remembers is that the missing digits were distinct, non-zero, and were in ascending order. Based on this information, she dials one number. What is the probability that she dials a wrong number?


[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

[Brent@GMATPrepNow]

Lola forgot last three digits of one telephone number. All she remembers is that the missing digits were distinct, non-zero, and were in ascending order. Based on this information, she dials one number. What is the probability that she dials a wrong number?
[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

Some examples of 3-digit numbers that meet the above criteria include 123, 156, 246, 489, 579, etc.

So, the question requires us to determine how many such numbers are possible.

Approach #1: Take the 9-digit number 123456789 and choose any 6 digits to eliminate.
Example: if we eliminate the digits 1,3,4,5,6 and 8, we're left with the 3-digit number 279
In how many ways can we select 6 digits to eliminate? Well, there are 9 digits altogether, and we must select 6. This can be accomplished in 9C6 ways (or 84 ways)

Approach #2: Take the 9 digits (1,2,3,4,5,6,7,8 and 9) and choose any 3.
Now list those 3 digits in ascending order, and you have a 3-digit number that meets the given criteria.
Note: there's only 1 way to list 3 unique digits in ascending order.
In how many ways can we select 3 digits? Well, there are 9 digits altogether, and we must select 3. This can be accomplished in 9C3 ways (or 84 ways)

Aside: Notice that 9C6 = 9C3 = 84. In general, nCr = nC(n-r)

Okay, so we now know that there are 84 different 3-digit numbers, and only 1 of those numbers is the correct number. In other words, 83 of the 84 numbers are the wrong number.

So . . . P(wrong number) = 83/84

Cheers,
Brent

[krusta80]

Lola forgot last three digits of one telephone number. All she remembers is that the missing digits were distinct, non-zero, and were in ascending order. Based on this information, she dials one number. What is the probability that she dials a wrong number?


[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

We can also enumerate the first few possibilities and extrapolate the answer like so...

assuming XYZ represents the three digits:

For X = 1, we have the following possibilities for Z based on each value for Y

Y,possible Z's
2,7 (3 through 9)
3,6 (4 through 9)
4,5 (5 through 9)
5,4 (6 through 9)
6,3 (7 through 9)
7,2 (8 through 9)
8,1 (9 through 9)

The total possibilities, therefore, for X=1 is the sum of 1 through 7, which equals 8C2.

Now, for X = 2, we can see how there would be one less possiblity for Z, when given each value of Y. So it'll be the sum of 1 through 6, which is 7C2.

I'm seeing a pattern here...which ends at the highest possible value for X,Y,Z: 7,8,9. So, the last value of X will give us 1 possibility, so we know when to stop -> at 2C2

Combining it all we get...

8C2+7C2+6C2+5C2+4C2+3C2+2C2 = 28+21+15+10+6+3+1 = 84 possible 3-digit sequences as described above

P(wrong one) = 83/84

[sanju09]

Thanks Brent and krusta80 for your decent inputs. Brent's way is marvelous and less complicated. krusta80's intelligent extrapolation method may sound 'rocket science' to many (minus me)

Thanks again

[krusta80]

Thanks Brent and krusta80 for your decent inputs. Brent's way is marvelous and less complicated. krusta80's intelligent extrapolation method may sound 'rocket science' to many (minus me)

Thanks again

I must say that I loved Brent's approach to this problem...very clever.

[rahul jaiswal]

The other way to solve this problem is to find out the probability of dialing last 3 digits wrong.
Like for last 3 digits in ascending order first no. should be in between (1 to 7), second no. should be in between (2 to 8) & third no. should be in between ( 3 to 9).
If Lola dials first no. either 8 or 9 then, she is dialing wrong no. for sure.
Similarly, if she dials second no. either 1 or 9 then, she is dialing wrong no. for sure.
Similarly, if she dials third no. either 1 or 2 then, she is dialing wrong no. for sure.

So, can't we work out probability of dialing wrong no. at first place, which would be 2/7. Proability of dialing wrong no. at second place, which would be 2/7 & similarly & probability of dialing wrong no. at third place, which would be again 2/7.

Please help me to figure out if this approach is right.

- regards,
Rahul

[Brent@GMATPrepNow]

The other way to solve this problem is to find out the probability of dialing last 3 digits wrong.
Like for last 3 digits in ascending order first no. should be in between (1 to 7), second no. should be in between (2 to 8) & third no. should be in between ( 3 to 9).
If Lola dials first no. either 8 or 9 then, she is dialing wrong no. for sure.
Similarly, if she dials second no. either 1 or 9 then, she is dialing wrong no. for sure.
Similarly, if she dials third no. either 1 or 2 then, she is dialing wrong no. for sure.

So, can't we work out probability of dialing wrong no. at first place, which would be 2/7. Proability of dialing wrong no. at second place, which would be 2/7 & similarly & probability of dialing wrong no. at third place, which would be again 2/7.

Please help me to figure out if this approach is right.

- regards,
Rahul

Hi Rahul,

This approach doesn't really address the question.

Lola already knows that the 3 digits are in ascending order. So, the digits that she enters will be in ascending order. Given that she is entering the digits in ascending order, our goal is to determine the probability that she does not enter the correct 3 digits.

Your solution suggests that she may not enter the 3 digits in ascending order.

Cheers,
Brent