6. We place in a bag 5 bills of 500 DA, 7 bills of 1000 DA and 10 bills of 2000 DA. A handful of 8 bills
are chosen at random, each bill having the same probability of being caught.
1. What is the probability of not having chosen any 500 DA banknotes?
2. What is the probability of having obtained only 2000 DA bills?
3. Repeat the experiment by drawing the bills one by one and putting the bill back in the bag after drawing it.
Calculate the probabilities of the two events above in this new experiment. TELL F THE ORDER MATER IN EACH AND WHY

Let's tackle each part of the problem step by step. This problem involves combinatorics and probability concepts.

### Total Bills and Initial Setup

We have three types of bills:
- 5 bills of 500 DA
- 7 bills of 1000 DA
- 10 bills of 2000 DA

The total number of bills is:
\[
5 + 7 + 10 = 22 \text{ bills}
\]
We are drawing a handful of 8 bills at random.

### **1. Probability of Not Choosing Any 500 DA Bills**

If we want to find the probability of not choosing any 500 DA bills, we need to consider only the 17 remaining bills (7 bills of 1000 DA and 10 bills of 2000 DA).

**Number of favorable outcomes:**
The number of ways to choose 8 bills from the 17 bills that are not 500 DA is given by the combination formula:
\[
\binom{17}{8}
\]

**Total number of outcomes:**
The total number of ways to choose 8 bills from all 22 bills is:
\[
\binom{22}{8}
\]

**Probability:**
The probability of not choosing any 500 DA bills is:
\[
P(\text{no 500 DA bills}) = \frac{\binom{17}{8}}{\binom{22}{8}}
\]

### **2. Probability of Obtaining Only 2000 DA Bills**

For this case, we want to choose all 8 bills from the 10 available 2000 DA bills.

**Number of favorable outcomes:**
The number of ways to choose 8 bills from the 10 bills of 2000 DA is:
\[
\binom{10}{8}
\]

**Total number of outcomes:**
Again, the total number of ways to choose 8 bills from all 22 bills is:
\[
\binom{22}{8}
\]

**Probability:**
The probability of obtaining only 2000 DA bills is:
\[
P(\text{only 2000 DA bills}) = \frac{\binom{10}{8}}{\binom{22}{8}}
\]

### **3. New Experiment: Drawing Bills One by One with Replacement**

When drawing bills one by one with replacement, the situation changes significantly. In this case, each draw is independent, and the probability of drawing each type of bill remains constant.

**Total Probability Calculation:**
- **Probability of drawing a 500 DA bill in any draw:** \( \frac{5}{22} \)
- **Probability of drawing a 1000 DA bill in any draw:** \( \frac{7}{22} \)
- **Probability of drawing a 2000 DA bill in any draw:** \( \frac{10}{22} \)

**(a) Probability of not choosing any 500 DA bills:**

To not choose any 500 DA bills in 8 independent draws:
\[
P(\text{no 500 DA bills in any draw}) = \left(1 - \frac{5}{22}\right)^8 = \left(\frac{17}{22}\right)^8
\]

**(b) Probability of obtaining only 2000 DA bills:**

To obtain only 2000 DA bills in 8 independent draws:
\[
P(\text{only 2000 DA bills}) = \left(\frac{10}{22}\right)^8
\]

### **Does Order Matter in Each Case?**

- **In the first two scenarios (without replacement):** The order does **not** matter because we're choosing 8 bills simultaneously, and the probability depends only on the combination of bills chosen.

- **In the new experiment (with replacement):** The order **does** matter because each draw is independent, and the sequence of draws affects the outcome. However, since we’re considering identical outcomes (e.g., no 500 DA bills or only 2000 DA bills), the specific sequence of outcomes isn’t crucial, but the repetition of the same type of event is.

Would you like a detailed calculation for these probabilities?

### **Related Questions:**

1. How would the probability change if you were to choose 9 bills instead of 8?
2. What is the probability of choosing exactly 4 bills of 1000 DA and 4 bills of 2000 DA?
3. How would the probabilities differ if there were 15 bills of 2000 DA instead of 10?
4. What is the expected number of 500 DA bills in a random selection of 8 bills?
5. How would the problem change if the bills were drawn without replacement but only one at a time?

**Tip:** In combinatorial problems, recognizing whether order matters (permutations) or doesn't (combinations) is crucial to solving them correctly.

Explore the probability of selecting bills from a bag containing different denominations (500 DA, 1000 DA, and 2000 DA) in various scenarios. Understand the calculation of probabilities for not choosing any 500 DA bills and obtaining only 2000 DA bills. Includes detailed combinatorics and probability formulas. Suitable for advanced high school students and enthusiasts of probability theory.

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