An assembly line has two inspectors. The probability that the first inspector will miss a defective item is 0.05. If the defective item passes the first inspector, the probability that the second inspector will miss it is 0.01. What is the probability of the event that a defective item will pass by both inspectors?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
0.005
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To find the probability that a defective item will pass by both inspectors, we need to multiply the probabilities of each inspector missing the defect.
Let's break down the problem step by step:
1. First, the probability that the first inspector will miss a defective item is given as 0.05. This means that the probability of the first inspector identifying a defective item is 1 - 0.05 = 0.95.
2. If the defective item passes the first inspector, we then move to the second inspector. The probability that the second inspector will miss the defect is given as 0.01, which means that the probability of the second inspector identifying a defective item is 1 - 0.01 = 0.99.
3. We want to find the probability that both inspectors will miss the defect, so we multiply the probabilities together: 0.95 * 0.99 = 0.9405.
Therefore, the probability that a defective item will pass by both inspectors is 0.9405, or 94.05%.