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This problem have been taken from the book' An Introduction to Probability Theory and Its Applications' by Williams Feller(1906-1970)

Note:- Assume in each case that all possible arrangements have the same probability.

Ten pair of shoes are in a closet. Four shoes are selected at random.Find the probability that there will be at least one pair among the four shoes selected

Solution:-

Answer provided by the author is $\frac{\binom{55}{2}}{\binom{20}{4}}=\frac{1485}{4845}=\frac{99}{323}$

we want to find the probability that there will be at least one pair of shoes among the four shoes selected which is equal to the probability that remains after deducting the probability of no pairs of shoes among the four shoes selected from the total probability. Let us calculate the probability of picking 1st,2nd,3rd and 4th shoes so that there are no pairs.

The probability of 1st shoes 20/20

2nd shoes 18/19, 3rd shoes 16/18 and 4th shoes 14/17. If we deduct the product of these result from the total probability, we get the our desired result.i-e $1-\frac{20*18*16*14}{20*19*18*17}=\frac{99}{323}$ combination

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    $\begingroup$ What is the rationale for your formula? If you can justify it using the axioms of probability, it's likely correct; but if you cannot find a justification, you have no basis to suppose it is right. (BTW, Feller's answer is correct.) $\endgroup$
    – whuber ♦
    Commented Sep 21, 2018 at 15:39

1 Answer 1

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Comment: Here is a simulation in R of a million draws of four shoes from the closet. In the closet each pair has its own number from 1 to 10. If my draw results in exactly four uniquely different numbers, I have no pairs. Otherwise, I have at least one pair. The vector nr.unique has a million entries, each of which can be a number from 2 (I drew 2 pairs of shoes) through 4 (no pairs).

The vector nr.uniq < 4 is a 'logical' vector containing TRUEs and FALSEs. Its mean is its proportion of TRUEs.

With a million draws the margin of simulation error should be less than 0.001 in 95% of such simulations. Results agree with Feller's answer.

set.seed(921)
closet = rep(1:10, 2)
nr.uniq = replicate( 10^6, length(unique(sample(closet,4))) )
mean(nr.uniq < 4);  99/323
[1] 0.306618      # simulated result
[1] 0.3065015     # Feller's answer

table(nr.uniq)    # tally counts
nr.uniq
     2      3      4 
  9377 297241 693382 
2*sd(nr.uniq < 4)/sqrt(10^6)
[1] 0.0009221792  # aprx 95% margin of sim err
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  • $\begingroup$ Question is asking probability of at least one pair of shoes in the selected four number of shoes.So this is equal to probability of one pair of shoes and probability of two pair of shoes in the selection of four shoes out of ten pairs of shoes. I have posted exactly the same answer.So my answer is correct.Author might have forgotten to consider the probability of success and failure. $\endgroup$
    – Win_odd Dhamnekar
    Commented Sep 22, 2018 at 2:45
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    $\begingroup$ Some misunderstanding here. Feller's answer is correct. Your view would have more weight if accompanied by a combinatorial argument in addition to claims of correctness. $\endgroup$
    – BruceET
    Commented Sep 22, 2018 at 3:05
  • $\begingroup$ Author had given correct answer. My answer was wrong. Now I have provided the correct answer by editing my question. $\endgroup$
    – Win_odd Dhamnekar
    Commented Sep 28, 2018 at 11:26

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