The partial probability distribution of X, the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given below. If one of those automobiled is equally likely to have either 2 or 3 defective tires, what is P[(X = 2) U (X = 4)] = P(2 U 4)?
x | 0 | 1 | 2 | 3 | 4 |
---|---|---|---|---|---|
P(x) | .54 | .12 | .20 |
The sum of all the probabilities should be 1. Hence,
P(0) + P(1) + P(2) + P(3) + P(4) = 1
0.54 + 0.12 + P(2) + P(3) + 0.20 = 1
P(2) + P(3) = 0.14
Since 2 and 3 are equally likely:
P(2) = P(3) = 0.14/2 = 0.07
Therefore,
P[(X = 2) U (X = 4)] = 0.07 + 0.20 = 0.27
The partial probability distribution of X, the number of defective tires on a randomly selected automobile checked at a...
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