Feller shoe-matching problem - Cross Validated most recent 30 from stats.stackexchange.com 2019-07-19T23:22:29Z https://stats.stackexchange.com/feeds/question/368036 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/368036 2 Feller shoe-matching problem Dhamnekar Winod https://stats.stackexchange.com/users/72126 2018-09-21T15:06:09Z 2018-09-30T19:44:30Z <p>This problem have been taken from the book' An Introduction to Probability Theory and Its Applications' by Williams Feller(1906-1970)</p> <p>Note:- Assume in each case that all possible arrangements have the same probability.</p> <p>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</p> <p>Solution:-</p> <p>Answer provided by the author is <span class="math-container">$\frac{\binom{55}{2}}{\binom{20}{4}}=\frac{1485}{4845}=\frac{99}{323}$</span></p> <p>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. </p> <p>The probability of 1st shoes 20/20</p> <p>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 <span class="math-container">$1-\frac{20*18*16*14}{20*19*18*17}=\frac{99}{323}$</span> <a href="http://math.stackexchange.com/questions/546493/probability-of-no-matching-or-exactly-one-matching-">combination</a></p> https://stats.stackexchange.com/questions/368036/-/368109#368109 3 Answer by BruceET for Feller shoe-matching problem BruceET https://stats.stackexchange.com/users/85665 2018-09-22T00:05:13Z 2018-09-22T00:49:54Z <p><strong>Comment:</strong> 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 <code>nr.unique</code> has a million entries, each of which can be a number from 2 (I drew 2 pairs of shoes) through 4 (no pairs).</p> <p>The vector <code>nr.uniq &lt; 4</code> is a 'logical' vector containing <code>TRUE</code>s and <code>FALSE</code>s. Its <code>mean</code> is its proportion of <code>TRUE</code>s.</p> <p>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.</p> <pre><code>set.seed(921) closet = rep(1:10, 2) nr.uniq = replicate( 10^6, length(unique(sample(closet,4))) ) mean(nr.uniq &lt; 4); 99/323  0.306618 # simulated result  0.3065015 # Feller's answer table(nr.uniq) # tally counts nr.uniq 2 3 4 9377 297241 693382 2*sd(nr.uniq &lt; 4)/sqrt(10^6)  0.0009221792 # aprx 95% margin of sim err </code></pre>