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There are 10 pairs of socks. What is the probability that in 4 socks chosen at random there is at least one pair.

My try: Let $A$ be an event of choosing exactly one pair of socks among 4 socks and $B$ be an event of choosing exactly two pairs,

$$P(A)=\frac{\binom{10}{1}\left(1-\frac{\binom{9}{1}}{\binom{18}{2}}\right)}{\binom{20}{4}}$$ and

$$P(B)=\frac{\binom{10}{2}}{\binom{20}{4}}$$

So the total probability is $P(A)+P(B)$.

But i know that some mistake is there in my solution... can any one help?

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  • $\begingroup$ if we do without compliment method..whats wrong in my method $\endgroup$ – Ekaveera Kumar Sharma Feb 19 '15 at 15:05
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Calculate $1$ minus the probability of the complementary event:

The number of ways to choose $4$ out of $20$ socks is:

  • Choose the $1$st sock out of $20$ socks
  • Choose the $2$nd sock out of $19$ socks
  • Choose the $3$rd sock out of $18$ socks
  • Choose the $4$th sock out of $17$ socks

The number of ways to choose $4$ out of $20$ socks with no pairs is:

  • Choose the $1$st sock out of $20$ socks
  • Choose the $2$nd sock out of $18$ socks
  • Choose the $3$rd sock out of $16$ socks
  • Choose the $4$th sock out of $14$ socks

So the probability of choosing $4$ out of $20$ socks with at least one pair is:

$$1-\frac{20\cdot18\cdot16\cdot14}{20\cdot19\cdot18\cdot17}$$


Please note that I've essentially taken into account the order of the socks.

If I chose not to take it into account, then I would need to divide each result by $4!$.

But since this factor appears in both the numerator and the denominator, I can ignore it.

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  • $\begingroup$ The OP did not ask for the correct calculation; he asked what is wrong with his own calculation. This does not answer the question. $\endgroup$ – WillO Feb 19 '15 at 15:05
  • $\begingroup$ @WillO: It does, however, propose an alternative solution, which is nevertheless a constructive contribution (and at least $4$ people thought the same). Perhaps OP shouldn't mark it as the accepted answer, but it still does not deserve to be down-voted. $\endgroup$ – barak manos Feb 19 '15 at 15:18
  • $\begingroup$ To make your answer clear: Let's say I'd like to know what is the probability that in $6$ six socks chosen randomly out of 10 pairs, there isn't any pair been chosen. that will be: $$\frac{20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot 10}{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15}$$ is that correct? $\endgroup$ – Jneven Nov 17 '18 at 16:40
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There is no pair of socks among the 4 socks picked if and only if the 4 socks belong to different pairs. The number of possibilities for this to happen is $2^4\cdot\binom{10}{4}$ (first, choose which 4 pairs out of 10 are going to be hit; then for each pair, which of the two socks is picked).

In total, how many choices exist for choosing 4 socks out of 20? $\binom{20}{4}$. So the probability to "miss" every single pair of socks is $$\frac{2^4\cdot\binom{10}{4}}{\binom{20}{4}}=\frac{224}{323}$$, and the quantitity you are looking for is $1-\frac{2^4\cdot\binom{10}{4}}{\binom{20}{4}} = \frac{99}{323}.$

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  • $\begingroup$ The OP did not ask for the correct calculation; he asked what is wrong with his own calculation. This does not answer the question. $\endgroup$ – WillO Feb 19 '15 at 15:05
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    $\begingroup$ @WillO: It does, however, propose an alternative solution, which is nevertheless a constructive contribution (and at least $2$ people thought the same). Perhaps OP shouldn't mark it as the accepted answer, but it still does not deserve to be down-voted. $\endgroup$ – barak manos Feb 19 '15 at 15:18
  • $\begingroup$ how can you find the probability for the case for which $4$ socks chosen randomly there is exactly two pairs? $\endgroup$ – Jneven Nov 17 '18 at 16:49
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Your numerator in $P(A)$ is wrong—it isn't even an integer! The numerator had to represent the number of ways that exactly one pair could be chosen:

$$^{10}\mathrm C_r\times\,^9\mathrm C_2\times2^2$$

Then you will get the same answer as CC: $\displaystyle\frac{99}{323}$

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This is late, but here is a general answer (studying for an exam, so I figured this was good practice).

There are $2n$ shoes ($n$ pairs). So the denominator is ${2n\choose 2r}$ ie, choosing $2r$ shoes from $2n$ total. We find our numerator by calculating how many ways we have no matching pairs. This means we would choose our shoes from the total pairs, and thus choose $2r$ from $n$, ${n\choose 2r}$. We then would choose ${2\choose 1}$ shoes, and multiply this result $2r$ times. This is because if we want to have no pairs, we would only select one shoe from every two to ensure there are no pairs chosen. ${n\choose 2r}$ different shoe types and $\underbrace{{2\choose 1}\cdot{2\choose 1}\cdot\ldots \cdot{2\choose 1}}_{2r \text{ of these}}$ ways of choosing one shoe from each pair ensuring no correct pairs. Distinguishing the chosen specified shoes leads to

$$\boxed{P(\text{no pairs})=\frac{{n\choose 2r}2^{2r}}{{2n\choose 2r}}}$$

If we want to find the probability of finding exactly one pair, then we note there are ${n\choose 1}$ ways to select one pair. Then we want to select $2r-2$ pairs from $n-1$ shoes that remain, mathematically we can express this as ${n-1\choose 2r-2}$. If we take one shoe from each pair, we would do this $2r-2$ times, so from the same reasoning as earlier, $$\boxed{P(\text{one pair})=\frac{n{n-1\choose 2r-2}2^{2r-2}}{{2n\choose 2r}}}$$

Generalizing to getting exactly $i$ pairs

$$\boxed{P(\text{exactly i pairs})=\frac{{n\choose i}{n-i\choose 2r-2i}2^{2r-2i}}{{2n\choose 2r}}}$$

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