Question

There is one error in one of five blocks of a program. To find the error, we test three randomly selected blocks. Let X be the number of errors in these three blocks. Compute E(X) and Var(X)

Answer 1

Probability(error in one block)= 1/5
Probability(error in 3 blocks)= 3*(1/5)= 0.6

Let, 
     0 = No error
     1 = error
Hence.
E(x) = 0.4 x 0 x+0.6 + 1 
        = 1.6

Answer 2

Answer with explanation:

Number of Blocks in the program =5

Number of errors in the Program =1

Probability of an event

            $=\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}$

Probability of getting an error in a program out of five programs

                               $=\frac{1}{5}$

Now, three programs are selected randomly

Required probability that is Expectation E(x)

        =1 ×Error in first one and No error in last two +2×Error in Second one and No error in first and last +3×Error in Third one and No error first and Second one

  $E(x=1,2,3)=1 \times \frac{1}{5}\times \frac{4}{5}\times \frac{4}{5}+2 \times \frac{4}{5}\times \frac{1}{5}\times \frac{4}{5}+3 \times \frac{4}{5}\times\frac{4}{5}\times \frac{1}{5}\\\\=\frac{16}{125}+\frac{32}{125}+\frac{48}{125}\\\\ =\frac{96}{125}$

→E(x²)

    $E(x=1^2,2^2,3^2)\\\\E(x=1,4,9)=1 \times \frac{1}{5}\times \frac{4}{5}\times \frac{4}{5}+4 \times \frac{4}{5}\times \frac{1}{5}\times \frac{4}{5}+9 \times \frac{4}{5}\times\frac{4}{5}\times \frac{1}{5}\\\\=\frac{16}{125}+\frac{64}{125}+\frac{144}{125}\\\\ =\frac{224}{125}$  

Variance (x)

      =E(x²)-E(x)

   $=\frac{224}{125}-\frac{96}{125}\\\\=\frac{128}{125}\\\\=1.024$