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svp [43]
3 years ago
7

A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two def

ectives are located. Once she locates the two defectives, she stops testing, but the second defective is tested to ensure accuracy. Let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y.
Mathematics
1 answer:
Stells [14]3 years ago
4 0

Answer:

when Y = 2, P(2) = 1/6 = 0.167

when Y = 3, P(3) = 2/6 = 0.333

when Y = 4, P(4) = 3/6 = 0.5

Step-by-step explanation:

Total number of possible ways to choose 2 components for defectives out of 4 components = 4C2 = 6

Y can only take values of 2, 3 and 4

So, Probability of Y = P(Y) = 1C(Y-1) / 6

when Y = 2, P(2) = 1/6 = 0.167

when Y = 3, P(3) = 2/6 = 0.333

when Y = 4, P(4) = 3/6 = 0.5

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