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A box contains 4 defective and 6 good machines. Two machines are selected at random without replacement. Find the probability that both the machines are good.
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Verified Answer
The correct answer is:
\(\frac{1}{3}\)
Box contains 4 defective and 6 good machines.
\(\therefore\) Total number of machines \(=10\)
Probability that first machine selected is good
\(=\frac{\text { Number of good } m / c}{\text { Total number of } m / c}=\frac{6}{10}\)
Probability that second machine is good \(=\frac{5}{9}\)
Probability that both machines are good
\(=\frac{6}{10} \times \frac{5}{9}=\frac{1}{3} .\)
\(\therefore\) Total number of machines \(=10\)
Probability that first machine selected is good
\(=\frac{\text { Number of good } m / c}{\text { Total number of } m / c}=\frac{6}{10}\)
Probability that second machine is good \(=\frac{5}{9}\)
Probability that both machines are good
\(=\frac{6}{10} \times \frac{5}{9}=\frac{1}{3} .\)
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