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Four children write their name on a piece of paper, put these pieces of paper in a pot and then draw one piece of it after having mashed up the pieces.

What is the probability that:

a) Each child draws a piece of paper where his own name is not on it b) Exactly one child draws a piece of paper with his name on it c) Two children draw a piece of paper with their name on it d) Three children draw a piece of paper with their name on it e) All children draw a piece of paper with their name on it

For e) I think the probability is 1/24, since we have in total 4*3*2*1 possibilities. d) sounds sort of impossible...I am not even sure what it means..if three children draw a piece of paper with their own name then actually all do. c) seems to be to be $\binom{4}{2}(\frac{1}{24})^{4}$. b) seems to me to be 4*$(\frac{1}{24})^{2}*\frac{3}{24}*\frac{2}{21}$ and a) $\frac{3}{24}*\frac{2}{24}*\frac{1}{24}$. Am I right? Thank you

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  • $\begingroup$ Try derangements. (d) and (e) are the same only when (d) means atleast 3 not exactly 3. $\endgroup$ – user67773 Mar 23 '13 at 9:01
  • $\begingroup$ I still dont see how "at least 3" and "exactly 3" differ in this case. And I dont know what you mean by derangements. $\endgroup$ – TestGuest Mar 23 '13 at 9:06
  • $\begingroup$ @TestGuest: You're unlikely to get a lot of help here if you're not willing to invest the minimal time to google terms you don't know. $\endgroup$ – joriki Mar 23 '13 at 10:01
  • $\begingroup$ I am sorry, but obviously I have already invested time in this exercise..it is not like I just posted a question with no effort from my side..and I understand derangement but imo thats what I have done already. What should I have answered? $\endgroup$ – TestGuest Mar 23 '13 at 10:02
  • $\begingroup$ If you understand "derangement", why did you write "I don't know what you mean by derangements"? $\endgroup$ – joriki Mar 23 '13 at 10:13
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Since you have only 4 children in problem and thus 24 possible outcomes, it is easy to write down all of them and see which one's satisfy conditions. Let us assign numbers to children and denote outcomes by the following way: $(a_1,a_2,a_3,a_4)$ will mean that child #1 picked a paper with number $a_1$, child #2 picked with $a_2$ and so on.

a) $(2,1,4,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,1,2), (3,4,2,1), (4,1,2,3), (4,3,1,2), (4,3,2,1)$

Only these outcomes satisfy condition a) so the answer will be $\frac{9}{24}$.

b) $(1,3,4,2),(1,4,2,3),(3,2,4,1),(4,2,1,3),(2,4,3,1),(4,1,3,2),(2,3,1,4),(3,1,2,4)$
c) $(1,2,4,3),(1,4,3,2),(1,3,2,4),(4,2,3,1),(3,2,1,4),(2,1,3,4)$
d) and e) $(1,2,3,4)$

Note that for arbitrary $n$ number of children there is no nice formula to express the answer.

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  • $\begingroup$ Thank you very much! Even though there is not a general formula for all the cases, is it not nevertheless possible to use standard combinatorics (as I tried) for each of the cases? And still I am not so sure about c), what am I not understanding? $\endgroup$ – TestGuest Mar 23 '13 at 10:30
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    $\begingroup$ I've edited the answer. d) and e) are the same. It is possible to use combinatorics, but you'll have to deal with conditional probabilities and probably end up with some sum, but you will not get a nice formula. $\endgroup$ – gev Mar 23 '13 at 10:42
  • $\begingroup$ Thank you so much! So for e) I get still 1/24 for d) I get 0, since my interpretation is that it is impossible (your interpretation is a possibility too though), for c) I get 8/24 and hence adding up together and subtracting the denominators from 24 for b) I get 6/24. Is this correct? $\endgroup$ – TestGuest Mar 23 '13 at 10:46
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    $\begingroup$ I read the question again and it seems d) is impossible, if we interpret it like c). So yes, you are right. $\endgroup$ – gev Mar 23 '13 at 10:53

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