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To find the joint probability mass function of \(N_1\) and \(N_2\), we need to determine the probability of each combination of their possible values. The probabilities can be calculated using combinatorics. Let's create a table representing the possible values of \(N_1\) in rows and the possible values of \(N_2\) in columns: \[ \begin{array}{c|c|c|c} & N_2 = 1 & N_2 = 2 & N_2 = 3 \\ \hline N_1 = 1 & P_{11} & P_{12} & P_{13} \\ \hline N_1 = 2 & P_{21} & P_{22} & P_{23} \\ \hline N_1 = 3 & P_{31} & P_{32} & P_{33} \\ \hline N_1 = 4 & P_{41} & P_{42} & P_{43} \\ \end{array} \] Now, let's calculate the probabilities for each combination of \(N_1\) and \(N_2\): - \(P_{11}\): In this case, the first transistor tested is defective, and the second one is also defective. The probability of this happening is \(\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\) - \(P_{12}\): In this case, the first transistor tested is defective, and the third one is defective. The probability of this happening is \(\frac{2}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{1}{10}\) - \(P_{13}\): In this case, the first transistor tested is defective, and the fourth or fifth one is defective. The probability of this happening is \(\frac{2}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{1}{5}\) - \(P_{21}\): In this case, the second transistor tested is defective, and the third one is also defective. The probability of this happening is \(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{1}{10}\) - \(P_{22}\): In this case, the second transistor tested is defective, and the fourth one is defective. The probability of this happening is \(\frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} = \frac{1}{5}\) - \(P_{23}\): This case is not possible, since after testing three transistors, there are only two left, and no additional test is needed. - \(P_{31}\): In this case, the third transistor tested is defective, and the fourth one is also defective. The probability of this happening is \(\frac{3}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{5}\) - \(P_{32}\): This case is not possible, since after testing four transistors, there is only one left, and no additional test is needed. - \(P_{33}\): This case is not possible, since after testing four transistors, there is only one left, and no additional test is needed. - \(P_{41}\): In this case, the fourth transistor tested is defective, and the fifth one is also defective. The probability of this happening is \(\frac{3}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{5}\) - \(P_{42}\): This case is not possible, since after testing all five transistors, no additional test is needed. - \(P_{43}\): This case is not possible, since after testing all five transistors, no additional test is needed.

Now that we have all the probabilities for each combination of \(N_1\) and \(N_2\), we can represent the joint probability mass function as: \[ P(N_1, N_2) = \begin{cases} \frac{1}{10} \ & \text{if} \ N_1 = 1 \ \text{and} \ N_2 = 1 \\ \frac{1}{10} \ & \text{if} \ N_1 = 1 \ \text{and} \ N_2 = 2 \\ \frac{1}{5} \ & \text{if} \ N_1 = 1 \ \text{and} \ N_2 = 3 \\ \frac{1}{10} \ & \text{if} \ N_1 = 2 \ \text{and} \ N_2 = 1 \\ \frac{1}{5} \ & \text{if} \ N_1 = 2 \ \text{and} \ N_2 = 2 \\ 0 \ & \text{if} \ N_1 = 2 \ \text{and} \ N_2 = 3 \\ \frac{1}{5} \ & \text{if} \ N_1 = 3 \ \text{and} \ N_2 = 1 \\ 0 \ & \text{if} \ N_1 = 3 \ \text{and} \ N_2 = 2 \\ 0 \ & \text{if} \ N_1 = 3 \ \text{and} \ N_2 = 3 \\ \frac{1}{5} \ & \text{if} \ N_1 = 4 \ \text{and} \ N_2 = 1 \\ 0 \ & \text{if} \ N_1 = 4 \ \text{and} \ N_2 = 2 \\ 0 \ & \text{if} \ N_1 = 4 \ \text{and} \ N_2 = 3 \\ \end{cases} \]