A lot of 50 items has 40 good items and 10 bad items.
1
(a) Suppose we test five samples from the lot, with replacement. Let X be the
number of defective items in the sample. Find P[X = k].
(b) Suppose we test five samples from the lot, without replacement. Let Y be the
number of defective items in the sample. Find P[Y = k].

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.

Explore the probability of finding defective items in a sample of 5 items from a lot of 50 items, with and without replacement. Understand the binomial and hypergeometric distributions used to calculate these probabilities.

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