An urn contains 10 pebbles numbered from 0 to 9. Three balls are drawn in succession from the urn with replacement. After each a draw a number is associated to the pebble and registered. If the numbers associated with the pebbles is for example 015 is drawn, we get $15.

(i) what is the probability of getting more than $9.

(ii) what is the probability of getting $100.

(iii) what is the probability of getting less than $100.

  • $\begingroup$ I don't think words like "tricky" belong in the title; that's a very subjective assessment. Also "combinatorics problem" is information that's conveyed in the tags; the title should be more specific; for instance "probabilities for numbers with uniformly random decimal digits". $\endgroup$ – joriki Jun 5 '12 at 4:09
  • $\begingroup$ @joriki edited appropriately... $\endgroup$ – count Jun 5 '12 at 4:10

Remember all values are divided by 10^3 to find out the probability. Because balls 1,2 and 3 have 0-9 possible values.

iii) is simple, 1st ball has only one possible value 0 and balls 2 and 3 have 0-9, which makes 10^2.

ii) number of ways of getting 100 is just 1.

i) number of possible entries of 1 and 2 is just 1 (00) and there are 10 possible entries for the last ball 0-9, which makes it 10 possible ways of getting a value less than or equal to 009, subtract this from 10^3 and that's your answer.

Divide all numbers by 10^3.

Please correct me if i've missed out on something.

  • $\begingroup$ a) I think you meant 1-9 instead of 1-10? b) You need to enclose $\TeX$ code like 10^3 in dollar signs to get it formatted: single dollar signs for inline formulas, double dollar signs for displayed equations. $\endgroup$ – joriki Jun 5 '12 at 4:15
  • $\begingroup$ umm.. sure.. changes noted. first time posting on here, so getting used to it.. and yea, that was a typo. I did mean 1-9. $\endgroup$ – Vigneshwaren Jun 5 '12 at 4:19
  • $\begingroup$ You can edit your post using the "edit" link underneath it. I think you should at least correct the typo, and preferably also improve the formatting. Note that you can use \cdot to get a proper multiplication dot instead of the asterisk. $\endgroup$ – joriki Jun 5 '12 at 4:37
  • $\begingroup$ @user1150428: I don't quite get (iii). this is what I think. to get less than $100, shouldn't the first pebble be 0, and then the second and third could be anything from 0-9? $\endgroup$ – count Jun 5 '12 at 4:42
  • $\begingroup$ I think you're right, count, but I also think it would have been better for you to include what you think, and why you think it, in your question. That would make it easier for people to see what, if anything, you don't understand, and to target answers to helping you with what you need. $\endgroup$ – Gerry Myerson Jun 5 '12 at 5:36

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