The problem describes an inspector testing four components, two of which are defective, and asks to find the probability distribution of $Y$, where $Y$ denotes the test on which the second defective component is found.

Steps to solve the problem:

1. Understand the scenario:

   • There are 4 components, and two are defective.
   • The inspector tests the components one by one until both defectives are found, testing the second defective to ensure accuracy.

2. Possible values of $Y$:

   • $Y$ can take values from 2 to 4, because:
     • The second defective cannot be found on the first test (as it's impossible to find both defectives on the first test).
     • The second defective must be found on the 2nd, 3rd, or 4th test.

3. Combinatorial Setup:

   • The inspector randomly selects the components to test. Since two of the four components are defective, the probability of finding the second defective depends on how the tests progress.

4. Probability Distribution for $Y$:

   • $Y = 2$: The second defective is found on the second test. This occurs if the inspector finds the first defective on the first test and the second defective on the second test.
   • $Y = 3$: The second defective is found on the third test. This means the first defective was found in either of the first two tests, and the second defective was found in the third.
   • $Y = 4$: The second defective is found on the fourth test.

5. Detailed Calculation:

   • The total number of ways to select 2 defectives from 4 components is $\binom{4}{2} = 6$.

   • For $Y = 2$: The first defective must be found in the first test and the second defective in the second test. There are $\binom{2}{1} = 2$ favorable outcomes (choosing 2 defectives out of 4 with the first defective appearing first), so the probability is $\frac{2}{6} = \frac{1}{3}$.

   • For $Y = 3$: The first defective is found in one of the first two tests, and the second defective in the third test. The number of favorable outcomes for this is $\binom{2}{1} \times \binom{2}{1} = 2 \times 1 = 2$, so the probability is $\frac{2}{6} = \frac{1}{3}$.

   • For $Y = 4$: The first defective is found in one of the first three tests, and the second defective is found in the fourth test. The number of favorable outcomes is $\binom{2}{1} \times 1 = 1$, so the probability is $\frac{1}{6}$.

6. Probability Distribution of $Y$: $P(Y = 2) = \frac{1}{3}, \quad P(Y = 3) = \frac{1}{3}, \quad P(Y = 4) = \frac{1}{6}.$

Would you like further clarifications or details?

Related Questions:

1. How does the number of components affect the probability distribution for finding defectives?
2. What if there were three defective components instead of two? How would the solution change?
3. How would the distribution change if the inspector stops testing after finding the first defective?
4. Could this problem be solved using a geometric distribution? Why or why not?
5. How does the order of testing affect the overall probability of finding the defectives early?

Tip:

Always consider all possible outcomes and the restrictions on values when determining probability distributions for similar inspection problems.