tag:blogger.com,1999:blog-5871184605628305545.post2888854308863167971..comments2013-05-25T03:36:33.414-07:00Comments on Puzzled: Repeated AlbhabetsAnonymoushttp://www.blogger.com/profile/02364548875097103884noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-5871184605628305545.post-43525400338786319952013-04-02T05:30:55.024-07:002013-04-02T05:30:55.024-07:00Let the letters a to z be a1,a2,....a26
Now, let a...Let the letters a to z be a1,a2,....a26<br />Now, let a1=1 if letter 'a' is repeated ow 0<br />Similarly for a2 to a26<br />Now, <br />Let N=number of alphabets repeated = a1+a2+...+a26<br />Then,<br />E(N)= E(a1)+E(a2)+....+E(a26)<br />Now,<br />E(a1) = 1*P(a1=1) +0*P(a1=0)<br />E(a1) = P(a1= 1) = 1-P(a1=0)<br />Now,<br />P(a1=0) = letter 'a' doesn't appear on the 36 word or appears only once<br />P(a1=0) = [36C0*(1/26)^0*(25/26)^36] + <br /> [36C1*(1/26)^1*(25/26)^35]<br /><br />Thus, we can find the expected alphabets that would be repeated.Anonymoushttps://www.blogger.com/profile/02364548875097103884noreply@blogger.com