A machine manufactured by a firm consists of two parts A and B. But of

1.	A machine manufactured by a firm consists of two parts A and B. But of 100 A’s manufactured, 9 are likely to be defective and out of 100 B’s manufactured 5 are likely to be defective. Find the probability that a machine manufactured by the firm is free of any defect. Give your answer, rounded off to two places of decimal.
Answer» Let E = {part A is defective}, F = {part B is defective}. Then, P (E) = \(\frac{9}{100}\), P(F) = \(\frac{5}{100}\) ∴ P(\(\bar{E}\)) = 1 - \(\frac{9}{100}\) = \(\frac{91}{100}\) and P(\(\bar{F}\)) = 1 - \(\frac{5}{100}\) = \(\frac{95}{100}\) Since E and F are independent, therefore, \(\bar{E}\) and \(\bar{F}\) and are also independent. Now P (none is defective) = P(\(\bar{E}\)) . P(\(\bar{F}\)) = \(\frac{91}{100}\) x \(\frac{95}{100}\) = 0.91 x 0.95 = 0.8645 = 0.86 to 2 d.p.

A machine manufactured by a firm consists of two parts A and B. But of 100 A’s manufactured, 9 are likely to be defective and out of 100 B’s manufactured 5 are likely to be defective. Find the probability that a machine manufactured by the firm is free of any defect. Give your answer, rounded off to two places of decimal.

Answer»

Let E = {part A is defective}, F = {part B is defective}.

Then, P (E) = \(\frac{9}{100}\), P(F) = \(\frac{5}{100}\)

∴ P(\(\bar{E}\)) = 1 - \(\frac{9}{100}\) = \(\frac{91}{100}\) and P(\(\bar{F}\)) = 1 - \(\frac{5}{100}\) = \(\frac{95}{100}\)

Since E and F are independent, therefore, \(\bar{E}\) and \(\bar{F}\) and are also independent.

Now P (none is defective) = P(\(\bar{E}\)) . P(\(\bar{F}\)) = \(\frac{91}{100}\) x \(\frac{95}{100}\)

= 0.91 x 0.95 = 0.8645 = 0.86 to 2 d.p.

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