@ 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:
Code:
22211 22231 22241 22251 22261 22271 22281 22291 22201
22213 22233 22243 22253 22263 22273 22283 22293 22203
22214 22234 22244 22254 22264 22274 22284 22294 22204
22215 22235 22245 22255 22265 22275 22285 22295 22205
22216 22236 22246 22256 22266 22276 22286 22296 22206
22217 22237 22247 22257 22267 22277 22287 22297 22207
22218 22238 22248 22258 22268 22278 22288 22298 22208
22219 22239 22249 22259 22269 22279 22289 22299 22209
22210 22230 22240 22250 22260 22270 22280 22290 22200
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