Let the letters a to z be a1,a2,....a26 Now, let a1=1 if letter 'a' is repeated ow 0 Similarly for a2 to a26 Now, Let N=number of alphabets repeated = a1+a2+...+a26 Then, E(N)= E(a1)+E(a2)+....+E(a26) Now, E(a1) = 1*P(a1=1) +0*P(a1=0) E(a1) = P(a1= 1) = 1-P(a1=0) Now, P(a1=0) = letter 'a' doesn't appear on the 36 word or appears only once P(a1=0) = [36C0*(1/26)^0*(25/26)^36] + [36C1*(1/26)^1*(25/26)^35]
Thus, we can find the expected alphabets that would be repeated.
Let the letters a to z be a1,a2,....a26
ReplyDeleteNow, let a1=1 if letter 'a' is repeated ow 0
Similarly for a2 to a26
Now,
Let N=number of alphabets repeated = a1+a2+...+a26
Then,
E(N)= E(a1)+E(a2)+....+E(a26)
Now,
E(a1) = 1*P(a1=1) +0*P(a1=0)
E(a1) = P(a1= 1) = 1-P(a1=0)
Now,
P(a1=0) = letter 'a' doesn't appear on the 36 word or appears only once
P(a1=0) = [36C0*(1/26)^0*(25/26)^36] +
[36C1*(1/26)^1*(25/26)^35]
Thus, we can find the expected alphabets that would be repeated.