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grandymaker [24]
3 years ago
14

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 n

umber of errors in these three blocks. Compute E(X) and Var(X)
Mathematics
2 answers:
marusya05 [52]3 years ago
6 0
Probability(error in one block)= 1/5
<span>Probability(error in 3 blocks)= 3*(1/5)= 0.6</span>

Let, 
     0 = No error
<span>     1 = error
</span>Hence.
E(x) = 0.4 x 0 x+0.6 + 1 
        = 1.6
bekas [8.4K]3 years ago
6 0

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

       

 

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