There are 10 pairs of socks in a cupboard from which 4 individual socks are picked at random.The probability that there is at least on pair is
A)195/323
B)99/323
C)198/323
D)185/323

• the first sock can be chosen in any way. so it can be taken in 20/20 ways. next sock can be taken in 18/19 ways as we cannot take the paired sock for the first sock. similarly the next sock can be drawn in 16/18 and next in 14/17 ways. thus the probability that no paired sock in taken is 18*16*14/19*18*17 .. so the probability that atleast one paired sock is chosen is 1-(18*16*14/19*18*17)=99/323
• 11 years agoHelpfull: Yes(20) No(2)
• I think that the probability of atleast one pair = (1 - probability of matching no pair)
  
  so I proceeded as follows -
  
  4 socks are drawn in 20 C 4 ways.
  
  therefore n(s) = 20C4
  
  Now let us say
  
  Ist Row - 1 2 3 4 5 6 7 8 9 10
  
  2nd Row 1' 2' 3' 4' 5' 6' 7' 8' 9' 10'
  
  1st socks can be selected in 20 C 1 let say 1 from Ist row.
  
  2nd sock must not be the 1' therefore it can selected from remaining 18 socks in
  18 C 1 ways = 18 let say it is any socks like 6.
  
  3rd sock must not be 1' as well as 6' therefore It can be drwn from the remaining 16 balls. let say it is 10.
  
  similarly 4th ball is drawn in 14 ways
  
  probability of not matching even a single pair is
  1 - (20 . 18 . 16 . 14 )/20C4.
  
  Please guide me If I am on right way or missing out something.
  
  Also the book answer for this is 99/323.
  
  http://mathhelpforum.com/statistics/81783-need-assistance-socks-problem.html
• 11 years agoHelpfull: Yes(9) No(1)
• required probability: 1 - ((20C1 * 18C1 * 16C1 * 14C1/4!)/20C4) = 99/323
• 9 years agoHelpfull: Yes(3) No(0)