1. ## shoes

There are 6 pair of shoes. We choose 4 pair of shoes. What is the probability that chosen shoes form a pair?

C(12,4) is the total number of events.

If we had chosen one shoe probability would be 1/11.

I don't know what to do with 2 pairs.

2. ## Re: shoes

We choose 1 from 6 pairs:1/6
we choose 1 from the remaining 11:1/11

3. ## Re: shoes

I don't understand the question. If you choose 4 pairs of shoes, there is a 100% chance that the chosen shoes form pairs because you only chose pairs. If, instead, you choose 8 random shoes, now there is a nonzero probability that you will get unmatched shoes. Please state the problem more carefully.

4. ## Re: shoes

W e chose 4 shoes not 4 pair of shoes. I made a typo.

5. ## Re: shoes

Originally Posted by kastamonu
W e chose 4 shoes not 4 pair of shoes. I made a typo.
$P[\text{at least 1 pair}] = 1 - P[\text{no pairs}]$

$P[\text{no pairs}] = \dfrac{10}{11}\dfrac{8}{10}\dfrac{6}{9} = \dfrac {16}{33}$

$P[\text{at least 1 pair}]=\dfrac{17}{33}$

6. ## Re: shoes

Our book says 1/33
You chose 1 shoe. The probability of chosing no pair is 10/11
You chose 3 rd shoe the probability of chosing no-pair is 8/10

1-10/11.8/10 seems ok to me.
How did you find 6/8?

7. ## Re: shoes

Originally Posted by kastamonu
Our book says 1/33
the answer I gave is for at least 1 pair

is the question exactly one pair?

8. ## Re: shoes

Among 6 pairs of shoes we chose 4 pair of shoes.What is the probability that chosen shoes will form 2 pair of shoes?

9. ## Re: shoes

Originally Posted by kastamonu
Among 6 pairs of shoes we chose 4 pair of shoes.What is the probability that chosen shoes will form 2 pair of shoes?
You have already been told that you cannot mean pair.
You must mean four shoes period!

10. ## Re: shoes

Originally Posted by kastamonu
Among 6 pairs of shoes we chose 4 pair of shoes.What is the probability that chosen shoes will form 2 pair of shoes?
choosing 2 pairs completely determines the 4 shoes.

There are $\dbinom{6}{2}$ ways to form 2 pairs from the 6.

$p = \dfrac{\dbinom{6}{2}}{\dbinom{12}{4}} = \dfrac{15}{495}=\dfrac{1}{33}$

11. ## Re: shoes

Originally Posted by romsek
$P[\text{at least 1 pair}] = 1 - P[\text{no pairs}]$
$P[\text{no pairs}] = \dfrac{10}{11}\dfrac{8}{10}\dfrac{6}{9} = \dfrac {16}{33}$
$P[\text{at least 1 pair}]=\dfrac{17}{33}$
Originally Posted by romsek
the answer I gave is for at least 1 pair
I just don't see it. I do not think you showed that. Please explain.

Originally Posted by kastamonu
Among 6 pairs of shoes we chose 4 pair of shoes.What is the probability that chosen shoes will form 2 pair of shoes?
$\begin{array}{*{20}{c}} \text{left}&\| & {{L_1}}&{{L_2}}&{{L_3}}&{{L_4}}&{{L_5}}&{{L_6}} \\ \text{right}&\| & {{R_1}}&{{R_2}}&{{R_3}}&{{R_4}}&{{R_5}}&{{R_6}} \end{array}$ there are $\dbinom{12}{4}=495$ ways to select a four element subset from that twelve element set.

Suppose we choose the subset $\{L_1,L_2,L_3,R_4\}$ has no two elements have the same subscript.
Thus we have not selected a pair. How many four element subsets do not have two elements with the same subscript?
Some of those contain all $L_k's$; some all $R_k's$ ; some contain three $L_k's$ and one $R_k$; some contain three $R_k's$ and one $L_k$; AND some contain two of each. Now some of those are counted the same way. The first two and the next two the same number of ways.

Subtract that from the total and you have the number of subsets of four that contain at least one pair.

12. ## Re: shoes

Originally Posted by Plato
I just don't see it. I do not think you showed that. Please explain.

$\begin{array}{*{20}{c}} \text{left}&\| & {{L_1}}&{{L_2}}&{{L_3}}&{{L_4}}&{{L_5}}&{{L_6}} \\ \text{right}&\| & {{R_1}}&{{R_2}}&{{R_3}}&{{R_4}}&{{R_5}}&{{R_6}} \end{array}$ there are $\dbinom{12}{4}=495$ ways to select a four element subset from that twelve element set.

Suppose we choose the subset $\{L_1,L_2,L_3,R_4\}$ has no two elements have the same subscript.
Thus we have not selected a pair. How many four element subsets do not have two elements with the same subscript?
Some of those contain all $L_k's$; some all $R_k's$ ; some contain three $L_k's$ and one $R_k$; some contain three $R_k's$ and one $L_k$; AND some contain two of each. Now some of those are counted the same way. The first two and the next two the same number of ways.

Subtract that from the total and you have the number of subsets of four that contain at least one pair.
at least 1 pair is the complement of no pairs

pick a shoe w/probability 1

pick one not of it's pair w/probability $\dfrac {10}{11}$

again $\dfrac{8}{10}$

and once more $\dfrac{6}{9}$

The product is the probability of picking 4 shoes, none making a full pair.

1 minus this is the probability of at least 1 pair.

13. ## Re: shoes

There are $\dbinom{12}{4}$ ways to choose 4 shoes from 12. For exactly one pair of shoes, choose three pairs of shoes. Among those three, choose one pair to be the pair. For the other two, choose one shoe from each pair:

$\dfrac{\dbinom{6}{3}\dbinom{3}{1}\dbinom{2}{2} \dbinom{2}{1}\dbinom{2}{1}}{\dbinom{12}{4}} = \dfrac{240}{495} = \dfrac{16}{33}$

Add to that the probability of choosing exactly 2 pairs $\dfrac{1}{33}$ and you get the probability of at least one pair: $\dfrac{17}{33}$ as romsek found.

14. ## Re: shoes

Ran 3 simulations (at least 1 pair), 1 million each:
515,242
515,121
515,187
Confirms 17/33 (.515151515....)

15. ## Re: shoes

Originally Posted by SlipEternal
There are $\dbinom{12}{4}$ ways to choose 4 shoes from 12. For exactly one pair of shoes, choose three pairs of shoes. Among those three, choose one pair to be the pair. For the other two, choose one shoe from each pair:
$\dfrac{\dbinom{6}{3}\dbinom{3}{1}\dbinom{2}{2} \dbinom{2}{1}\dbinom{2}{1}}{\dbinom{12}{4}} = \dfrac{240}{495} = \dfrac{16}{33}$
Add to that the probability of choosing exactly 2 pairs $\dfrac{1}{33}$ and you get the probability of at least one pair: $\dfrac{17}{33}$ as romsek found.
$\begin{array}{*{20}{c}} \text{left}&\| & {{L_1}}&{{L_2}}&{{L_3}}&{{L_4}}&{{L_5}}&{{L_6}} \\ \text{right}&\| & {{R_1}}&{{R_2}}&{{R_3}}&{{R_4}}&{{R_5}}&{{R_6}} \end{array}$ there are $\dbinom{12}{4}=495$ ways to select a four element subset from that twelve element set. Suppose we choose the subset $\{L_1,L_2,L_3,R_4\}$ has no two elements have the same subscript.
There are thirty ways to have four element subsets either all right or all left. There 120 4-subsets that are three of one and one of the other. There 90 4-subsets that are two left and two right. In each case no two elements have the same subscript. Therefore there are 240 ways to select four of these shoes getting no pair. Thus there are 255 ways to get at least one pair,
$\dfrac{255}{495}=0.515151515151$. We all agree.
Look at this totally different approach.