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Three identical books were randomly tidied up in a cabinet containing five shelves, the probability that three books are on the same shelf is:

My thoughts:

I think the answer is $\displaystyle {1 \over 25}$

indeed,

  • Alternative solution ( Combinatorics ).

    The contingencies are the functions from the set of all books $\mathcal{L}=\{L_{1},L_{2},L_{3}\}$ onto the set of all shelves $\mathcal{E}=\{E_{1},E_{2},...,E_{5}\}$, there are $5^3$ such functions. These contingencies are equally likely. Among these possibilities, there are $5$ which put the books on the same shelf.

  • Alternative solution ( Probability )

    The probability that book $i$ either on the shelf with number $j$ is $\dfrac{1}{5}$. The probability that all three books are on the shelf number $j$ is therefore : $\left(\dfrac{1}{5}\right)^{3}$. The probability that they are all on the same shelf is: $5\times \left(\dfrac{1}{5}\right)^{3}$.

  • Is my proof correct ? I'm also interested in other ways to solve it

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No matter where we place the first book, the probability each of the other two is on the same shelf as the first book is $\frac{1}{5}$. I agree, the probability is $\frac{1}{25}$.

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  • $\begingroup$ Thanks for fixing my english $\endgroup$ – Educ Jan 17 '16 at 18:21
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    $\begingroup$ No problem, happy to help! $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 18:28
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Your answer is correct, assuming that each of the books are placed independently, and each shelf is equally likely.

However, if the phrase "randomly tidied up" instead meant "choose a random arrangement of books", then you get a different answer.

  • There are $5$ ways where they are all on the same shelf.
  • There are $5\cdot 4=20$ ways where there are two on shelf, and the other on a different shelf.
  • Three are $\binom{5}{3}=10$ ways where they are all on different shelves.

Assuming each of these arrangements is equally likely, the desired probability is $\frac{5}{35}=\frac17$.

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  • $\begingroup$ where did you come up with $35$ $\endgroup$ – Educ Jan 17 '16 at 18:50
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    $\begingroup$ $5+10+20{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 18:52
  • $\begingroup$ @dREaM Thanks again $\endgroup$ – Educ Jan 17 '16 at 18:53

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