A drawer contains eight different pairs of socks. If six socks are taken at random without replacement, compute the probability that there is at least one matching pair among these six socks. Hint: compute the probability that there is not a matching pair first.

Sorry guys, I'm really awful at this class, I'm trying to train my brain to think the right way to solve these on my own but it's hard.


Ok my suggestion is the following.

You surely know how to use the hint: If you denote the probability of not having a matching term by $P$ then the probability of having at least one matching pair is ... (I'll leave it to you to think about this one)

Now you might want to use the following formula.

Let $\Omega$ be a (finite) probability space and $A\subset \Omega$ an event, then for the probability of $A$ holds$$ P(A)=\frac{|A|}{|\Omega|} $$

That means the probability of $A$ is the number of all results such that $A$ has happened divided by the total number of possible results.

Now you have to compute those cardinalities: $\binom{n}{k}$ denotes the number of possibilities to draw $k$ items out of $n$ items when you do not care about the order in which the items were drawn (which is the case in your problem).

So $|\Omega|=\binom{16}{6}$ in your case, because you draw 6 socks out of 16 socks in total.

Furthermore there are $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ possibilities to draw $6$ socks without having a matching pair when taking into account the order (why?). Furthermore you have $6!$ possibilities to arrange $6$ objects on $6$ slots, so you have to divide by that number. I'll leave the rest to you. I hope this was helpful (and I didn't make a mistake somewhere).

EDIT: I just read your comment asking to explain the $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ part. Well this is just counting socks. I'll call the socks that result in the event "no matching pairs" as good socks and any sock preventing that result a bad sock.

  • In the first round you cannot get a matching pair obviuosly (as you'd need $2$ socks for a pair) so the number of good socks is 16.
  • In the second round you have only 15 socks left, however one of those is of the same color as the one you already have and therefore a bad sock. Makes in total 15-1=14 good socks
  • round three: you have 14 socks left. Among those are 2 bad socks (the corresponding sock to your first and second pick) so you have 14-2=12 socks left.
  • and so on.

You also asked for a way to write this with binomials. Look at the following$$ \frac{\binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{0} \cdot \binom{2}{0} \cdot \binom{8}{2}}{\binom{16}{6}} $$

What do I want to tell you with that fraction? Well we now consider pairs. One pair consists of $2$ socks and we draw $6$ socks in total. That means we want exactly one sock out of $6$ pairs (therefore $6$ times $\binom{2}{1}$) and no sock out of the resulting $2$ pairs (therefore the two $\binom{2}{0}$). Of course you could just omit the $\binom{2}{0}$ term as it is $1$ anyway, I just put it there for illustration. Ok and now you have $\binom{8}{2}$ possibilities to choose "which pair you don't get any sock from" ( I hope that sounds remotely logical).

If you type it in a calculator you'll find that the two results are the same.

  • $\begingroup$ Can you explain the 16*14*12*10*8*6? $\endgroup$ – Cole Howie Sep 3 '14 at 14:57
  • 1
    $\begingroup$ You have to normalise the $16\cdot 14 \cdot 12\cdot 10\cdot 8\cdot 6$ by $6!$, as otherwise you are taking into account the order. $\endgroup$ – snar Sep 3 '14 at 16:03
  • $\begingroup$ I knew i forgot something... you're of course right, thanks for the comment! $\endgroup$ – Barkas Sep 4 '14 at 7:19

The hint should help. You can always think about probability from the opposite event, though it's not always simpler. One primary rule of probability is to remember:

$P(A) + P(\lnot A) = 1$

So to find $P(A)$ you can always calculate $1-P(\lnot A)$.

Here, the probability of at least one matching pair of socks is complicated, because we could have 1 matching pair of any of the colors and 4 non-matching socks. We could have 2 matching pairs of any two of the colors and 2 non-matching socks. OR we could have 3 matching pairs of 3 of the colors. Whenever we see many options like this (often typified in a problem by the phrases "at least" or "at most"), that's a clue that calculating $1-P(\lnot A)$ is something to consider.

So, in this case $P(\lnot A)$ is the probability that we have NO matching socks of the 6 that we drew. Let's look at each sock.

First sock: it doesn't matter what sock we draw, since we have nothing to match it to yet. Probability of getting a sock that's acceptable = 1. 15 socks left.

Second sock: Well, now we have a sock that we don't want to match, so we want one of the other 7 colors. That eliminates 1 sock from the desirable outcomes, so $P = \frac {14}{15}$ and we have 14 socks left in the drawer.

Third sock: Now two colors are out of contention, so there are two socks in the remaining pile that we don't want. $P = \frac {12}{14}$

Continue this same pattern until the sixth sock. Then multiply all the probabilities together, since you are drawing these socks out in series. Your end result will be:

P(no matching pairs) = $1 \cdot \frac {14}{15} \cdot \frac{12}{14} \cdot \frac{10}{13} \cdot \frac8{12} \cdot \frac6{11} = \frac{10 \cdot 8 \cdot 6}{15 \cdot 13 \cdot 11} = \frac{2 \cdot 8 \cdot 2}{13 \cdot 11} = \frac{32}{143}$

Take that result and subtract it from 1 to get your final answer! P(at least one matching pair) = $\frac {111}{143}$

  • $\begingroup$ It make sense, but do you think there is a way using binomials? It makes sense as I said, I just don't know how to think the right way to where if I had this problem I could do it on my own.I feel like theres no set way of doing these problems the same and that is why I get thrown off on these problems $\endgroup$ – Cole Howie Sep 3 '14 at 14:54

First a reformulation of your question to a more general case. I am working with balls having the same color instead of socks that belong to the same pair. Suppose that the drawer contains $nk$ balls. The balls are coloured and exactly $n$ of them have color $C_i$ for $i=1,2,\cdots,k$. If you take at random $m\leq k$ balls out then what is the probability that each ball has a different color?

The answer is: $$\binom{k}{m}n^m\binom{nk}{m}^{-1}$$

I will try to explain why. Taking $m$ balls out of $nk$ can be done on $\binom{nk}{m}$ ways. If it is done under the restriction that each ball must have a different colour, then from the $k$ colours $m$ must be elected and this can be done on $\binom{k}{m}$ ways. For a specific selected color a ball must be chosen and this can be done on $n$ ways. This is to be done for any selected color so leads to $n^m$ ways. So under the restriction there are $\binom{k}{m}n^m$ ways.

In your case we have $k=8$, $n=2$, $m=6$ leading to a probability of $\binom{8}{6}2^6\binom{16}{6}^{-1}=\frac{32}{143}$ of taking $6$ socks (=balls) that do not match (=have different color). Then of course $\frac{111}{143}$ is the probability of taking $6$ socks such that there is at least one matching pair.

  • $\begingroup$ You have a minor typo - in paragraph 2 you write $n^k$ but it is $n^m$ elsewhere. $\endgroup$ – Duncan Sep 3 '14 at 17:11
  • $\begingroup$ @Duncan Thank you! I repaired. $\endgroup$ – drhab Sep 3 '14 at 19:50


Number of ways you can choose 6 socks when there are no pairs in them $A = 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$

Total Number of ways you can choose 6 socks out of 8 pairs of socks B= $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$

Probability that you will not find a match in the 6 drawn $= \frac{A}{B}$

Probability that you will find atleast one pair of socks in the 6 drawn $= 1-\frac{A}{B}$


The first sock could be chosen in 16 ways, the second sock could only be chosen in $14$ ways, the third sockk could be chosen in 12 ways ( each time, you are deducting one that has already been taken and the pair sock that goes along with that). Thus the numerator is $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$. For the denominator, you do not have any restriction, so, you pick the first sock in $16$ ways and $2^{\text{nd}}$ sock in $15$ ways, $3^{\text{rd}}$ sock in $14$ ways and until you pick the $6^{\text{th}}$th sock in $11$ ways. Thus it is $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$.




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