91Ó°ÊÓ

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.

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 \(P(x=0)=0.3\), \(P(x=1)=0.6\), \(P(x=2)=0.1\)