Answer:
144
Step-by-step explanation:
 
        
                    
             
        
        
        
The expected length of code for one encoded symbol is

where  is the probability of picking the letter
 is the probability of picking the letter  , and
, and  is the length of code needed to encode
 is the length of code needed to encode  .
.  is given to us, and we have
 is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length  , we would then expect
, we would then expect ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n) .
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length  , we would get
, we would get ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n) .
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114/2=57 I believe 57 is the correct answer
        
             
        
        
        
Answer:
 
 
Step-by-step explanation:
Complement of the event 

 
        
             
        
        
        
You can do like this
This is my answer 1 to 10
Do it now