This question has been answered here:

There are 5 pairs of shoes out of which 4 shoes are taken one by one. What is the probability that at least one pair is present in the shoes taken?

As have similar questions for example here:

Probability of having at least one pair by drawing 4 shoes from 12 pairs.

I understand the solutions above, what I don't understand is the solution given in the book:

The method they use is the same as above calculate one minus the probability of there being no pairs present.

$$1-\frac{\binom{5}{0}\binom{5}{4} + \binom{5}{1}\binom{4}{3} +\binom{5}{2}\binom{3}{2} +\binom{5}{3}\binom{2}{1}+ \binom{5}{4}\binom{1}{0}}{\binom{10}{4}}$$

My initial interpretation is that in the numerator:

1.) The first binomial term is the number of pairs present.
2.) The second binomial is the number of ways to chose the "other shoes" so that the total chosen is four.

However, it would appear that my interpretation of these two binomial terms must be incorrect?

  1. If I were correct about the first binomial term what would $\binom{5}{0}$ represent? All the shoes must come from one of the pairs so it's not possible to have none of the pairs represented?

  2. Also what these "other shoes" be in the second binomial term? Again every shoe is chosen from the 5 pairs so the 5 pairs are present every time?

Can someone please give me the correct interpretation of the solution in the book?


  • 3
    $\begingroup$ I would peronally prefer to do the calculation differently, by picking one shoe at a time and looking at the probability of not getting a pair from that shoe: $1-\frac{10}{10}\cdot \frac{8}{9}\cdot\frac{6}{8}\cdot \frac{4}{7}$. Seems like less work. $\endgroup$ – Arthur Sep 28 '18 at 8:42
  • $\begingroup$ Agreed I find that solution to be simpler but I just wanted to understand the solution in the book $\endgroup$ – Bazman Sep 28 '18 at 8:46

If we number the ten shoes $a_1a_2, b_1b_2, c_1c_2, d_1d_2, e_1e_2$, we can seperate them as follows:

$A = \{a_1, b_1, c_1, d_1, e_1\}$ and $B = \{a_2, b_2, c_2, d_2, e_2\}$

  • Then we can choose no shoe from $A$ and $4$ shoes from $B$ with $\binom{5}{0}\binom{5}{4}$.

  • We can choose $1$ shoe from $A$, say $i_1$, and in order not to have a pair, we can choose $3$ shoes from $B$ but except $i_1$ with $\binom{5}{1}\binom{4}{3}$

  • We can choose $2$ shoes from $A$, say $i_1$ and $j_1$, and in order not to have a pair, we can choose $2$ shoes from $B$ but except $i_1$ and $j_1$ with $\binom{5}{2}\binom{3}{2}$

Rest of them has the same interpretation so I leave them to you.


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