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Components coming off an assembly line are either free of defects (S, for success) or defective (F, for failure). Suppose that 80% of all such components are defect-free. Components are independently selected and tested one by one. Let y denote the number of components that must be tested until a defect-free component is obtained.
(a) What is the smallest possible y value, and what experimental outcome gives this y value? What is the second smallest y value, and what outcome gives rise to it?
The smallest value of y is ______ . The outcome corresponding to this is______ .
The second smallest value of y is_________ . The outcome that corresponds to this is ______ .
(b) Determine the probability of each of the five smallest y values. You should see a pattern that leads to a simple formula for P(y), the probability distribution of y.
P(y = 1) =
P(y = 2) =
P(y = 3) =
P(y = 4) =
P(y = 5) =

Sagot :

  1. The smallest possible value of y is 1, and the outcome corresponding to this is that the first component tested is defect-free. The second smallest value of y is 2, and the outcome corresponding to this is that the first component tested is defective, but the second component tested is defect-free.
  2. The probability of each of the five smallest y value is shown below.
  • P(y = 1) =  0.8
  • P(y = 2) = 0.16
  • P(y = 3) = 0.032
  • P(y = 4) = 0.0064
  • P(y = 5) = 0.00128

Probability can be used to make predictions or decisions in a variety of situations, such as in gambling, finance, and science. In these situations, probabilities can be calculated based on statistical data or by using mathematical models.

To determine the probability of each of the five smallest y values, we can use the probability of each possible outcome and the number of ways that outcome can occur.

P(y = 1) = probability that the first component is defect-free = 0.8

P(y = 2) = probability that the first component is defective and the second component is defect-free = (1 - 0.8) * 0.8 = 0.16

P(y = 3) = probability that the first two components are defective and the third component is defect-free = (1 - 0.8) * (1 - 0.8) * 0.8 = 0.032

P(y = 4) = probability that the first three components are defective and the fourth component is defect-free = (1 - 0.8) * (1 - 0.8) * (1 - 0.8) * 0.8 = 0.0064

P(y = 5) = probability that the first four components are defective and the fifth component is defect-free = (1 - 0.8) * (1 - 0.8) * (1 - 0.8) * (1 - 0.8) * 0.8 = 0.00128

We can see that the probability of each y value is simply the probability of the previous y value multiplied by (1 - 0.8) = 0.2. This means that the probability distribution of y is given by the formula P(y) = 0.2^(y - 1) * 0.8.

Learn more about probability, here brainly.com/question/11234923

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