AP EAMCET · Maths · Probability
If a toy factory, the machines \(A, B\) and \(C\) are used to manufacture \(30 \%, 40 \%\) and \(30 \%\) of the output, respectively. The probabilities of toys made by machines \(A, B\) and \(C\) to be defective are respectively \(2 \%, 3 \%\) and \(1 \%\). A toy is taken from the factory and is found to be defective. The probability that it was manufactured by the machine \(B\) is
- A \(4 / 5\)
- B \(2 / 9\)
- C \(3 / 4\)
- D \(4 / 7\)
Answer & Solution
Correct Answer
(D) \(4 / 7\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & P(A)=\frac{30}{100}, P(B)=\frac{40}{100}, P(C)=\frac{30}{100} \\ & P\left(\frac{E}{A}\right)=\frac{2}{100}, P\left(\frac{E}{B}\right)=\frac{3}{100}, P\left(\frac{E}{C}\right)=\frac{1}{100}\end{aligned}\) where \(E\) denotes defective toys. \(\therefore\)…
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