sanju09 wrote: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?
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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

rahul jaiswal wrote: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