Question 701874
<font face="Times New Roman" size="+2">


The conceptually simple way to do this is to calculate the probability that NONE of the 20 have the same last 5 digits and then subtract from 1.  The probability of "at least one" is the complement of "none".


The arithmetic is, on the other hand, god-awful ugly.


00000 to 99999 is 100,000 numbers.


The probability that in a group of 1 persons that the last five will NOT be duplicated is *[tex \LARGE \frac{100,000}{100,000}\ =\ 1].


The probability that in a group of 2 people that the last five will NOT be duplicated is *[tex \LARGE 1\ \cdot\ \frac{99,999}{100,000}]


Then for three: *[tex \LARGE 1\ \cdot\ \frac{99,999}{100,000}\ \cdot\ \frac{99,998}{100,000}]


And so on.  The entire calculation is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\Large{100,000}\LARGE P_{20}}{100,000^{20}}]


to get the probability that there are no pairs of students with the same last 5, and then subtract from 1 to get the probability that at least two people share the same last 5.  Happy calculator button pushing. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>