In a gross (144) of transistors, there are 2 defective ones. Three (3) transistors are selected...

In a gross (144) of transistors, there are 2 defective ones. Three (3) transistors are selected randomly. a. what is the number of ways one can select exactly one defective transistor among the three? b. what is the probability of selecting exactly one defective transistor among the three?

Expert Solution

a) Number of ways = 142C2 * 2C1 = 20022

b) Probability = (142C2 * 2C1)/144C3 = 0.0411

orchestra answered 2 years ago

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In a gross (144) of transistors, there are 2 defective ones. Three (3) transistors are selected...

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