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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
a. The ten different possible outcomes are: \((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\). b. The probability distribution of \(x\) is: \(\{0: 0.3, 1: 0.6, 2: 0.1\}\).

Step by step solution

01

Listing the possible outcomes

There are five boards and two are selected. Since order doesn't matter, this is a combination problem. Numbers from 1 to 5 are symbols standing for each board. We need to consider all possible pairs. They are: \((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\).
02

Determine the outcomes when x is defined

We define \(x\) to be the number of defective boards observed. Since Boards 1 and 2 are defective, the possible outcomes are: \(x = 0\) when pairs don't include 1 or 2 (namely (3, 4), (3, 5), (4, 5)), \(x = 1\) when pairs include only one of this defective boards (namely (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)), and \(x = 2\) when the pair is (1, 2).
03

Find the probability distribution

To find the probability of each value of x, we count the number of outcomes where x takes the particular value and divide it by the total possible outcomes. For \(x = 0\), there are 3 outcomes where \(x = 0\) out of a total of 10 possible outcomes. So, the probability is \(\frac{3}{10} = 0.3\). Similarly, for \(x = 1\), there are 6 outcomes where \(x = 1\) out of a total of 10 possible outcomes. So, the probability is \(\frac{6}{10} = 0.6\). And for \(x = 2\), there is 1 outcome where \(x = 2\) out of a total of 10 possible outcomes. So, the probability is \(\frac{1}{10} = 0.1\).

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Most popular questions from this chapter

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