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Chapter 7: Problem 12

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(x\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

Short Answer

Expert verified
Possible combinations are: \((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\). The probability distribution of selecting defective boards is: \(p(x=0) = 0.6, p(x=1) = 0.4, p(x=2) = 0.1\).

Step by step solution

01

Determine the Possible Outcomes

The possible number of ways to pick 2 boards out of 5 is represented by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n = 5 \) (the total number of boards) and \( r = 2 \) (the number of boards selected). Replacing into the formula gives \( C(5,2) = \frac{5!}{2!(5-2)!} = 10 \), which confirms that there are 10 possible outcomes.
02

List All Possible Outcomes

The possible outcomes of selecting 2 boards out of 5 can be listed as follows: \((1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)\).
03

Define Random Variable x

Let \(x\) be the number of defective boards observed among those inspected. In this context, \(x\) can take the values 0, 1 or 2 which represent no defective boards observed, one defective board observed, and two defective boards observed, respectively.
04

Construct the Probability Distribution for x

Now let's calculate the probabilities for each possible value of \(x\). Probability of \(x = 0\): This is the probability that none of the chosen boards are defective which corresponds to the outcomes \((1,3), (1,4), (1,5), (2,3), (2,4), (2,5)\), thus the probability is 6/10 = 0.6. Probability of \(x = 1\): This is the probability that exactly one of the chosen boards is defective which corresponds to the outcomes \((1,2), (3,4), (3,5), (4,5)\), thus the probability is 4/10 = 0.4. Probability of \(x = 2\): This is the probability that both chosen boards are defective which corresponds to the outcome \((1,2)\), thus the probability is 1/10 = 0.1.

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