A machine has 9 identical components that function independently. The probability that a component will fail is 0.2. The machine will stop working if 1 or 2 components fail. Find the probability that the machine will be fail?
Either-or probabilities are found by adding the individual probabilities.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(1 fail) = .2 * .98^8 = ?
P(2 fail) = .2^2 * .98^7 = ?
Sorry, I gave you the answer to not failing. Subtract the answer to above from 1.00.
Ignore my second post.
To find the probability that the machine will fail, we need to consider two cases: when one component fails and when two components fail.
Case 1: One component fails
The probability that one component fails is 0.2. Since there are 9 identical components, we need to choose one component to fail out of the 9. This can be done in 9C1 ways (9 choose 1), which is equal to 9. Therefore, the probability that one component fails is 0.2 * 9 = 1.8.
Case 2: Two components fail
The probability that two components fail is (0.2)^2 = 0.04. Similarly, we need to choose two components to fail out of the 9. This can be done in 9C2 ways (9 choose 2), which is equal to (9 * 8) / (2 * 1) = 36. Therefore, the probability that two components fail is 0.04 * 36 = 1.44.
To find the total probability of the machine failing, we need to add the probabilities from both cases:
Total probability = Probability of one component failing + Probability of two components failing
Total probability = 1.8 + 1.44
Total probability = 3.24
Therefore, the probability that the machine will fail is 3.24.