Q79 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

A computer program has two modules $X$ and $Y$ and errors in them occur independently. $X$ has an error with probability 0.1 and $Y$ has an error with probability 0.3 . If an error in $X$ alone cause the program to crash with probability 0.5 , an error in $Y$ alone causes the program to crash with probability 0.7 and an error in both $X$ and $Y$ cause the program to crash with probability 0.8 , then the probability that the program is crashed is

  • A
    $\frac{23}{125}$
  • B
    $\frac{26}{125}$
  • C
    $\frac{29}{125}$
  • Correct answer
    D
    $\frac{31}{125}$

Solution

Let the probability having error in $X$ is $P\left(E_1\right)=0.1$, the probability having error in $Y$ is $P\left(E_2\right)=0.3$ $\because$ Events are independent. $$ \therefore P\left(E_1 \cap E_2\right)=P\left(E_1\right) P\left(E_2\right)=0.03 $$ Now, probability having error in $X$ only is $$ (0.1-0.03)=0.07 $$ Probability having error in $Y$ only is $$ (0.3-0.03)=0.27 $$ It is given that program is crashed, if an error in $X$ causes alone is 0.5 , if error in $Y$ causes alone is 0.7 , and if error in both $X$ and $Y$ causes is 0.8 . So, required probability $$ \begin{aligned} & =(0.07 \times 0.5)+(0.27 \times 0.7)+(0.03 \times 0.8) \\ & =\frac{35+189+24}{1000}=\frac{248}{1000}=\frac{31}{125} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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