TS EAMCET · Maths · Probability
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 \(\frac{29}{125}\)
- D \(\frac{31}{125}\)
Answer & Solution
Correct Answer
(D) \(\frac{31}{125}\)
Step-by-step Solution
Detailed explanation
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…
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