Question 278589: Ten pair of shoes are kept in a rack.If four shoes are selected at random what is the probability that there is atleast one pair among them
Found 5 solutions by stanbon, edjones, Grinnell, geekyengineer, ikleyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Ten pair of shoes are kept in a rack.If four shoes are selected at random what is the probability that there is at least one pair among them
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There are ten shoes that do not contain a pair.
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P(at least one pair) = 1-P(no pair)
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P(no pair in the shoes selected) = 10C4/20C4 = 210/4845
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Therefore P(at least one pair) = 1 - 210/4845 = 4635/4845 = 0.9567
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Cheers,
Stan H.
Answer by edjones(8007)  (Show Source):
You can put this solution on YOUR website! After the 1st shoe removed 1/19 that its brother will be chosen. after a second removed then 2/18 chance and after 3rd chosen 3/17.
1/19 + 2/18 + 3/17 = .34
I am not 100% sure that this correct.
Let me know.
.
Ed
Answer by Grinnell(63) (Show Source):
You can put this solution on YOUR website! Another deceptive problem!!!!!
Out of the 20 shoes there are ten pairs.
Since we are trying to choose pairs,
4 selections of shoes are the same as 2
selections (attempts) at 1 pair.
Two (pair selections) picks out a 1/10 chance
at a pair would be 20%!
If anyone comes up with anything different, let me know.
Have not cracked prob. books in a while.
Thanks!
Answer by geekyengineer(2)  (Show Source):
You can put this solution on YOUR website! If four shoes selected at random the probability of at least one pair will be given by = 1 - p(no pair will be selected)
p(no pair will be selected) = (10C4 * 2^4)/(20C4) = 0.6934
Hence, required probabilty = 1 - 0.6934 = 0.3065
Answer by ikleyn(52756)  (Show Source):
You can put this solution on YOUR website! .
Ten pair of shoes are kept in a rack.If four shoes are selected at random what is the probability that there is atleast one pair among them
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Solution
Let's assume that the shoes are numbered by integer numbers from 1 to 20 in a way that
the first pair is (1,2),
the second pair is (3,4),
the third pair is (4,5),
. . . and so on . . .
the tenth pair is (19,20).
The full space of samples is the space of quadruples of different shoes taken from 20 at a time.
(it is the same as the set of all subsets consisting of 4 elements of the original set of 20 shoes).
So, the total number of elements of this sample space is  =  = 4845.
Very good.
Now I want to figure out, in how many ways I can choose 4 shoes in a way that there is NO pair/pairs among them.
Obviously, for it, I should choose these 4 shoes from DIFFERENT PAIRS.
In this way, I get the number of such quadruples as  =  = 3360.
The factor  is due to the fact that there is the choice of one of two shoes inside each pair.
So, the probability under the question is  =  =  = 0.3065 = 30.65%. ANSWER
Solved.
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My solution produces the same answer as the other tutor provides;
but I hope it gives more clear explanation on how the solution was obtained.
With it, my congratulations to tutor @geekyengineer, who found a brilliant solution !
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