Let's tackle both parts of the problem step by step.

Part (a): Sampling with Replacement

Given:

• Total items: 50
• Good items: 40
• Bad items (defective): 10
• Number of samples taken: 5 (with replacement)
• Let $X$ be the number of defective items in the sample.

Probability Mass Function (PMF) for $X$:

Since we are sampling with replacement, each sample is independent. The probability of picking a defective item in each trial is $\frac{10}{50} = \frac{1}{5}$, and the probability of picking a good item is $\frac{4}{5}$.

$X$ follows a Binomial distribution because the trials are independent, and each trial has the same probability of success (finding a defective item).

The probability mass function for a Binomial distribution is given by:

$P[X = k] = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 5$ (number of samples)
• $k$ is the number of defective items we want to find in the sample.
• $p = \frac{1}{5}$ (probability of finding a defective item)

So,

$P[X = k] = \binom{5}{k} \left(\frac{1}{5}\right)^k \left(\frac{4}{5}\right)^{5-k}$

Part (b): Sampling without Replacement

Given:

• Total items: 50
• Good items: 40
• Bad items (defective): 10
• Number of samples taken: 5 (without replacement)
• Let $Y$ be the number of defective items in the sample.

Probability Mass Function (PMF) for $Y$:

Since we are sampling without replacement, the trials are dependent, and we use the Hypergeometric distribution.

The probability mass function for a Hypergeometric distribution is given by:

$P[Y = k] = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}$

Where:

• $N = 50$ (total number of items)
• $D = 10$ (number of defective items)
• $n = 5$ (number of samples)
• $k$ is the number of defective items we want to find in the sample.

So,

$P[Y = k] = \frac{\binom{10}{k} \binom{40}{5-k}}{\binom{50}{5}}$

This formula gives the probability of finding exactly $k$ defective items in the sample when sampling without replacement.

Summary

• Part (a): $P[X = k] = \binom{5}{k} \left(\frac{1}{5}\right)^k \left(\frac{4}{5}\right)^{5-k}$
• Part (b): $P[Y = k] = \frac{\binom{10}{k} \binom{40}{5-k}}{\binom{50}{5}}$

Would you like further details or explanations on these solutions?

Follow-up Questions:

1. How does the probability change if the number of samples increases to 10 in part (a)?
2. What is the expected value of $X$ in part (a)?
3. What is the expected value of $Y$ in part (b)?
4. How do the variances of $X$ and $Y$ compare in the two scenarios?
5. How would the probability distribution change if there were only 5 defective items in the lot?

Tip:

In problems involving probability distributions, it's crucial to identify whether the sampling is done with or without replacement, as this determines whether to use the binomial or hypergeometric distribution.