1) A drawer contains n=5 different and distinguishable pairs of socks
(a total of 10 socks).  If a person randomly selects 4 socks, what is
the probability that there is no matching pair in the sample?

2) A drawer contains n different and distinguishable pairs of socks (a
total of 2n socks).  A person randomly selects 2r of the socks, where
2r > n.  In terms of n and r, what is the probability that there is no
matching pair in the sample?


I'm totally clueless on how to go about doing this, if you could
explain each step, I'd greatly appreciate it!

Thanks

Hello Dime365,

For #1 (choosing four socks out of ten)

Let's do the sequence of choices to explain the answer. I'll call the
pairs 1, 2, 3, 4, and 5 and the socks left (L) or right (R). So the
socks in the drawer are:
  1R  1L
  2R  2L
  3R  3L
  4R  4L
  5R  5L
Now choose a sock. Let's say you get 3R. You will have one sock at
this point and no pair with probability 1.

The only way to get a pair after selecting the second sock is to get
3L, a probability of 1*(1/9). You don't have a pair 8/9ths of the
time. Cumulative probability of a pair after choosing two socks is
1/9.

Assuming you don't have a pair, let's say you got 2L as the second
sock. Probability of this kind of outcome is 8/9. Now choose the third
sock; there are two possibilities (out of eight) of getting a pair (3L
and 2R). So the probability of getting a match with the third sock is
(8/9)*(2/8) or 2/9. Cumulative probability of a pair after choosing
three socks is 1/9+2/9 = 3/9 (or 1/3).

Assuming you don't have a pair, let's say you got 5R as the third
sock. Probability of this kind of outcome is 6/9 (or 2/3). Now choose
the fourth sock; there are three possibilities (out of seven) of
getting a pair (3L, 2R, and 5L). So the probability of getting a match
with the fourth sock is (2/3)*(3/7) or 2/7. Cumulative probability of
a pair after choosing four socks is 1/3+2/7; multiply the first
fraction by 7/7 and the second by 3/3 to get a common denominator or
7/21+6/21; the sum of that is 13/21 (roughly 61.905%).

In doing this answer, I walked through the sequence of steps of sock selection and
 - computed the probability of choosing a pair at each step
 - computed the cumulative probability of having a pair at each step
This was repeated for each selection until I had the correct answer.
This is a sequence of multiple events. For a brief overview of
multiple events, see
  http://www.800score.com/gre-guidec8bview1b.html
for the calculation for the probability of the "and" and "or" events.

For #2 (choosing 2r socks, 2r>n).

Perhaps this is a trick question, but the probability there is no
matching pair is ZERO. Let's do a simple example.

If n=3 (total 6 socks), and r=2 (so 2r=4); let's call the socks 
  GR  GL
  RR  RL
  BR  BL
to represent the colors red, green, and blue and left/right socks. If
you happened to pick three socks that did not match (e.g., GR, RR,
BR), by picking the fourth, you HAVE to get a matching pair. It does
not matter the size of "n" and "r" (as long as 2r>n). Another phrase
for this situation is certain probability.

There are a number of good sites that explain probability. I found a
few good ones using phrases such as
  probability pair of socks
  probability formula
  probability independent formula
  probability dependent formula

Let me know if you need a further explanation of the answer to your
question. I would be glad to respond to a clarification request.

  --Maniac

thanks!

To clarify, the answer to question 1, as asked, is 8/21 = .381