Question

O Probability Probability of independent events: Decimal answers At a factory that produces pistons for cars, Machine 1 produced 207 satisfactory pistons and 93 unsatisfactory pistons today. Machine 2 produced 279 satisfactory pistons and 31 unsatisfactory pistons today. Suppose that one piston from Machine 1 and one piston from Machine 2 are chosen at random from today's batch. What is the probability that the piston chosen from Machine 1 is satisfactory and the piston chosen from Machine 2 is unsatisfactory? Do not round your answer. (If necessary, consult a list of formulas.) Explanation Check Type here to search X 2023 McGew Hill LLC. All Rights Reserved. Terms of t 0/5 OI Cindy CADENA​

Answer 1

To find the probability that one piston from Machine 1 is satisfactory and one from Machine 2 is unsatisfactory, one must multiply the individual probabilities of these events occurring separately, resulting in a probability of 0.0221.

Step-by-step explanation:

To calculate the probability that the piston chosen from Machine 1 is satisfactory and the piston chosen from Machine 2 is unsatisfactory, we will apply the product rule of probability. This rule states that the probability of two independent events occurring together is the product of the individual probabilities of each event occurring alone.

For Machine 1, there are a total of 300 pistons (207 satisfactory + 93 unsatisfactory). Therefore, the probability that a piston chosen at random is satisfactory is:

P(Satisfactory from Machine 1) = Number of satisfactory pistons from Machine 1 / Total number of pistons from Machine 1 = 207/300.

For Machine 2, there are a total of 310 pistons (279 satisfactory + 31 unsatisfactory). Therefore, the probability that a piston chosen at random is unsatisfactory is:

P(Unsatisfactory from Machine 2) = Number of unsatisfactory pistons from Machine 2 / Total number of pistons from Machine 2 = 31/310.

Now, applying the product rule:

P(Satisfactory from Machine 1 AND Unsatisfactory from Machine 2) = P(Satisfactory from Machine 1) × P(Unsatisfactory from Machine 2) = (207/300) × (31/310)

By calculating we get:

P(Satisfactory from Machine 1 AND Unsatisfactory from Machine 2) = 207/300 × 31/310 = 0.0221

So, the probability is 0.0221 without rounding.