Question

at a factory that produces pistons for cars, machine 1 produced 693 satisfactory pistons and 297 unsatisfactory pistons today. machine 2 produced 880 satisfactory pistons and 220 unsatisfactory pistons today. suppose that one piston from machine 1 and one piston from machine 2 are chosen at random from todays batch. what is the probability that the piston chosen from machine 1 is unsatisfactory and the piston chosen from machine 2 is satisfactory? do not round your answer. (if necessary, consult a list of formulas.)

Explanation:

Step1: Calculate the probability of choosing an unsatisfactory piston from Machine 1

The total number of pistons from Machine 1 is \(693 + 297=990\).
The probability \(P(M1 - unsatisfactory)\) is \(\frac{297}{990}\).

Step2: Calculate the probability of choosing a satisfactory piston from Machine 2

The total number of pistons from Machine 2 is \(880+220 = 1100\).
The probability \(P(M2 - satisfactory)\) is \(\frac{880}{1100}\).

Step3: Use the formula for independent - event probability

Since the two events (choosing from Machine 1 and choosing from Machine 2) are independent, the probability \(P\) that the piston chosen from Machine 1 is unsatisfactory and the piston chosen from Machine 2 is satisfactory is \(P=P(M1 - unsatisfactory)\times P(M2 - satisfactory)\).
Substitute the values: \(P=\frac{297}{990}\times\frac{880}{1100}\).
Simplify \(\frac{297}{990}=\frac{297\div99}{990\div99}=\frac{3}{10}\), and \(\frac{880}{1100}=\frac{880\div220}{1100\div220}=\frac{4}{5}\).
Then \(P=\frac{3}{10}\times\frac{4}{5}=\frac{12}{50}= 0.24\).

Answer:

\(0.24\)