A box contains 4 defective and 6 good machines. Two machines are selected at random without replacement.…
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.
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{5}\)
Solution
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
\(=\frac{6}{10} \times \frac{5}{9}=\frac{1}{3} .\)