Question

QuestionLetS={1,4,8,16,32,64}be a sample space. IfP(1)=132andP(2k)=21−k for 2≤k≤6,find the expected value of the eventE={1,8,32,64}.Give your answer as a fraction in its simplest form.Provide your answer below:

Answer 1

In general, the expected value of an event X is given by the formula below

`E\lbrack X\rbrack=\sum_ix_i*f(x_i)`

Therefore, in our case,

`E\lbrack E\rbrack=1*P(1)+8*P(8)+32P(32)+64P(64)`

Thus,

`\begin{gathered} \Rightarrow E\lbrack E\rbrack=1*(1)/(32)+8*P(2^3)+32P(2^5)+64P(2^6) \\ \Rightarrow E\lbrack E\rbrack=(1)/(32)+8*2^(-2)+32*2^(-4)+64*2^(-5) \\ \Rightarrow E\lbrack E\rbrack=(1)/(32)+8*(1)/(4)+32*(1)/(16)+64*(1)/(32) \end{gathered}`

Then,

`\Rightarrow E\lbrack E\rbrack=(1)/(32)+2+2+2=6+(1)/(32)=(193)/(32)`

The answer is 193/32