Found this question elsewhere and think it was solved incorrectly. Give it a try. Only my solution to follow as no OA.

In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

(A) 0.81%
(B) 1%
(C) 1.53%
(D) 0.081%
(E) 1.44%


Additional question: Find the probability that any three digits (exactly three) are the same.
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First: there is some problems in calculations, then the answer doesn't match with the choices provided.
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Couldn't understand this part:


What exactly do you mean by double counting. Request you to explain this in detail.... Probably would be better if you could use some numbers as example and explain.

Thanks,
JT
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JT...........
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~~Better Burn Out... Than Fade Away~~

@ acegre:

OK I am still a bit lost as to why you do the substraction. Let me tell you my understanding. Might be u can see where am I leading to.

Lets revist the question again:
In a 5 digit ID number, what is the probability of exactly three digits are the digit 2?

Approach:
As rightly done by we could have 5C3 ways of constructing such number i.e. 10 ways.

1. 222XX
2. 22XX2
3. 2XX22
4. 2X2X2
5. 22X2X
6. 2X22X

7. XX222
8. X222X
9. X2X22
10.X22X2

Note: Although you came up with 10 ways, your listed 10 ways had an issue. Please check with the combination in ur post.

Considering the first 6 ways, which have 2 in as the first start digit (from left):
In 222XX, the first X from left can be filled up in 9 different ways (total 10 digits and we exclude digit 2 from this. Hence 9 digits)
Similarly the second X from left can also be filled up in 9 different ways. Reason Both X can be same also as there isn't any condition on it.
Therefore 222XX can be filled up in 9X9 ways i.e. 81. (I guess u got his number with a different approach which is also correct. I have used the generic approach followed in probability! But no worries as we both are on the same page ultimately)

Now comes the substraction which I do not understand the reason for! To make my point clear, I would list the 222XX combinations for you:



In the above table I dont see any number which is repeated or duplicated as per you. Hence this is where I am lost!

Anyway.. continuing the above approach... in the first 6 ways, we can have 81 X 6 = 486 numbers!

For the remaining 4 ways, we have the following explanation:
7. XX222
8. X222X
9. X2X22
10.X22X2

In the above sequence of numbers, the first X from left can be filled up by 8 different digits. (Out of the 10 digits, we remove 2 and 0 leaving only 8 possible digits)
The second X can anyway be filled up by 9 different digits (Leaving only 2 out of the 10 digits)

Hence for each way we have 8X9 = 72 numbers. Therefore for 4 ways we have 72X4 = 288 numbers!
Summing the 10 ways, we get 486 + 288 = 774 numbers
Total number possible = 9 X 10 X 10 X 10 X 10 = 90000 numbers

Therefore probability % = (774 / 90000) X 100 = 0.86%. I know the answer isn't among the options and this makes me feel a bit less confident. Do let me know if you identify any loops in my approach!

For the additional question stated:
Find the probability that any three digits (exactly three) are the same.
the approach is as follows:

Let us consider the following combinations. The combination would be same as consider for 2. Only difference is that we can replace 2 with Y for now.

Hence the combinations are:
1. YYYXX
2. YYXXY
3. YXXYY
4. YXYXY
5. YYXYX
6. YXYYX

7. XXYYY
8. XYYYX
9. XYXYY
10.XYYXY

Considering the first 6 ways:
YYYXX
- We can choose first Y in 9 different ways (Leave out 0 from 10 digits). Once first Y is selected, the other Ys would be the same and hence no selection required. The first X can be selected in 9 different ways (leave out the digit selected for Y from 10 digits). The second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits).
Therefore we have YYYXX as 9 x 1 x 1 x 9 x 8 = 648 numbers
Hence for 6 ways we have 648 x 6 = 3888 numbers

Considering the next 4 ways:
XXYYY
- We can chose the first Y in 10 different ways. As before, once first Y is selected, the other Ys would be the same and hence no selection required. Returning to X, the first X can be filled in 8 ways (leave out 0 and Y digit from 10 digits) and the second X can be selected in 8 ways (leave out the digit selected for Y and for the first X from 10 digits).
Therefore we have XXYYY as 8 x 8 x 10 x 1 x 1 = 880 numbers
Hence for the 4 ways we have 880 x 4 = 3520 numbers
Summing the numbers possible in 10 ways = 3888 + 3520 = 7408
Total Numbers = 90000 (as before)

Therefore probability = 7408/90000 = 0.0823 = 8.23%

Please feel free to share your analysis...

Thanks,
JT
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Cheers!
JT...........
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~~Better Burn Out... Than Fade Away~~

Chetan, I guess this is wat you r missing. Please check my explanation below:

First of all we would need to consider total number of 5 digit IDs as : 9 x 10 x 10 x 10 x 10 = 90000 and not 10^5 as we cannot have 0 in the first place, else it would become 4 digit ID.
Now as you consider for 3 places to be filled with digit 2, the possible number of combinations = 1 x 1 x 1 x 9 x 9 x 5c3 = 810 number. Again this includes the possibility of having digit 0 as the first or having digit 0 as both first and second. We need to subtract the number falling into this category from 810.
Hence we again arrive at the 4 combinations as follows where 0 could be the first or second:
XX222
X222X
X2X22
X22X2
Its only in them we can have 0 as first. If u notice, we could have 0 as both first and second in the first combination only.
For 0 as the first digit, we have the following posb = 1 x 1 x 1 x 1 x 9 x 4 = 36
For 0 as both as first and second digit we have only 1 posb i.e. 00222 or 222.
Therefore we need to subtract 36+1 = 37 from 810.
This gives us the actuall posb = 810-37 = 773

Hence probability % = \frac{773}{90000} x 100 = 0.858% \approx 0.86%

I know I do not match the options given but am a little lost as I feel the logic to be correct! I hope am on the right track!

Thanks,
JT
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JT...........
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~~Better Burn Out... Than Fade Away~~

I was beginning to rethink whether I should restart my Math prep again

Thanks for clarifying the same Bunuel!

Cheers!
JT
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Cheers!
JT...........
If u like my post..... payback in Kudos!!

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~~Better Burn Out... Than Fade Away~~

My approach was finding total possibilities, and dividing out the number of possibilities with exactly three 2s.

Total possibilities = 100000

Two of the digits have 9 possible choices each, and the other three must be 2s, and the number of ways of choosing the three spaces that must have 2s is 5c3 = 10

Desired outcomes = 9*9*1*1*1*10

Probability = 810/100000 = 0.81%

Hopefully this is correct!
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My GMAT quest...

...over!

Answer to the additional question:

favorable outcomes: 9C1.9C1.5C3.10C1/=8100
Total possible outcomes: 10^5

Probability in %= 8.1%

1x1x1x9x9 = 81. This much is understood. I also understand that there are 5C3 ways in which the number 2 can be distributed among the 5 places. But what happens when the other 2 places are filled by the same number? For example, if the ID has the digits 22255 then these can form 5!/3!x2! numbers i.e. 10 numbers. But if the ID has 22237, for example, then there are 5!/3! = 20 combinations. Therefore, we cannot simply perform the multiplication of 81 with 5C3.
In fact, we have to treat these cases separately as follows:
1x1x1x9x8 x 5!/3! = 1440 (here the other two numbers are different)
1x1x1x9x1 x 5!/3!x2!= 90 (here the other two numbers are the same)
Therefore, the total possibilities are 1440+90=1530.
Hence the prob. is 1530/90,000= .017%

I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5-step method:

1) Lay out the number of events (5 digits) _ _ _ _ _

2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)

3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5

4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10

5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081%