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 $\displaystyle {1 \over 25}$


  • 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


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 $\frac{1}{25}$.

  • $\begingroup$ Thanks for fixing my english $\endgroup$ – Educ Jan 17 '16 at 18:21
  • 1
    $\begingroup$ No problem, happy to help! $\endgroup$ – Jorge Fernández Hidalgo Jan 17 '16 at 18:28

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 $\frac{5}{35}=\frac17$.

  • $\begingroup$ where did you come up with $35$ $\endgroup$ – Educ Jan 17 '16 at 18:50
  • 2
    $\begingroup$ $5+10+20{}{}{}{}{}$ $\endgroup$ – 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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