A device can fail in four different ways with probabilities 1 = 0.2, 2 = 0.1,

A device can fail in four different ways with probabilities π1 = 0.2, π2 = 0.1, π³ = 0.4, and π4 = 0.3. Suppose there are 12 devices that fail independently of one another. What is the probability of 3 failures of the first kind, 4 of the second, 3 of the third, and 2 of the fourth?

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Probability of 3 failure of first type 02 Pro... View the full answer