Question

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. You 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. Let \(x=\) the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

Short answer

The 10 possible outcomes of selecting two boards from five 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 \(x\), the number of defective boards observed among those inspected, is \(P(x=0)=0.3\), \(P(x=1)=0.6\), \(P(x=2)=0.1\).

Step-by-step solution

List the 10 Different Possible Outcomes

The possible outcomes of selecting two boards out of five can be listed using combinations. In a combination, the order of selection does not matter. Hence, the possible pairs for selection are: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), and (4,5) total 10 combinations.

Calculate Probability Distribution of Inspecting Defective Boards

Let's denote \(x=\) the number of defective boards observed among those inspected. To find probability distribution of \(x\), we must consider all possible outcomes of \(x\), which are \(x=0\), \(x=1\), and \(x=2\). For each possible value of \(x\), we must calculate the corresponding probability. Probabilities can be calculated as follows: \n For \(x=0\), we can choose 0 defective boards and 2 non-defective boards from the lots. The total number of ways to do this is \(\binom{2}{0}*\binom{3}{2}=3\). So, the probability \(P(x=0) = \frac{3}{10}\) \n For \(x=1\), we can choose 1 defective board and 1 non-defective board from the lots. The total number of ways to do this is \(\binom{2}{1}*\binom{3}{1}=6\). So, the probability \(P(x=1) = \frac{6}{10}\) \n For \(x=2\), we can choose 2 defective boards and 0 non-defective boards from the lots. The total number of ways to do this is \(\binom{2}{2}*\binom{3}{0}=1\). So, the probability \(P(x=2) = \frac{1}{10}\) \n Hence, the probability distribution is \(P(x=0)=0.3\), \(P(x=1)=0.6\), \(P(x=2)=0.1\)