The probability of obtaining a raw chicken sandwich at Windies restaurant is 8.4%. Complete the table below to find the probability that the first raw chicken sandwich you get will be on the xth sandwich you buy.

Report answers accurate to 3 decimal places or copy paste from Excel.

xth sandwich bought	Probability this will be your first raw chicken sandwich
1	
2	
3	
4	
5	
6 or greater	


This problem deals with a **geometric distribution** since we are looking for the probability that the first success (getting a raw chicken sandwich) occurs on the xth trial (sandwich bought). The probability of success on each trial is given as \( p = 0.084 \), and the probability of failure is \( q = 1 - p = 0.916 \).

The probability that the first success occurs on the xth trial is given by the formula:

\[
P(X = x) = q^{x-1} \times p
\]

Where:
- \( p \) is the probability of success (getting a raw chicken sandwich).
- \( q \) is the probability of failure (not getting a raw chicken sandwich).
- \( x \) is the number of trials (sandwiches bought) until the first success.

### Let's calculate the probabilities:

1. **For x = 1**:
   \[
   P(X = 1) = (0.916)^{1-1} \times 0.084 = 1 \times 0.084 = 0.084
   \]

2. **For x = 2**:
   \[
   P(X = 2) = (0.916)^{2-1} \times 0.084 = 0.916 \times 0.084 = 0.076944
   \]
   Rounded to 3 decimal places: \( 0.077 \)

3. **For x = 3**:
   \[
   P(X = 3) = (0.916)^{3-1} \times 0.084 = 0.916^2 \times 0.084 = 0.839056 \times 0.084 = 0.070481
   \]
   Rounded to 3 decimal places: \( 0.070 \)

4. **For x = 4**:
   \[
   P(X = 4) = (0.916)^{4-1} \times 0.084 = 0.916^3 \times 0.084 = 0.768314 \times 0.084 = 0.06454
   \]
   Rounded to 3 decimal places: \( 0.065 \)

5. **For x = 5**:
   \[
   P(X = 5) = (0.916)^{5-1} \times 0.084 = 0.703646 \times 0.084 = 0.059106
   \]
   Rounded to 3 decimal places: \( 0.059 \)

6. **For x ≥ 6**:
   To find the cumulative probability that the first raw chicken sandwich will be on the 6th or later trial, we sum the probabilities for all values greater than or equal to 6:
   
   \[
   P(X \geq 6) = 1 - (P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))
   \]
   \[
   P(X \geq 6) = 1 - (0.084 + 0.077 + 0.070 + 0.065 + 0.059) = 1 - 0.355 = 0.645
   \]

### Completed table:

| xth sandwich bought | Probability this will be your first raw chicken sandwich |
|---------------------|----------------------------------------------------------|
| 1                   | 0.084                                                    |
| 2                   | 0.077                                                    |
| 3                   | 0.070                                                    |
| 4                   | 0.065                                                    |
| 5                   | 0.059                                                    |
| 6 or greater        | 0.645                                                    |

Would you like a more detailed breakdown of any step or a different question explored?

Here are 5 follow-up questions for further understanding:
1. What is the probability that the first raw chicken sandwich occurs within the first 3 sandwiches bought?
2. How does the probability change as the number of trials increases?
3. How would the probabilities change if the chance of getting a raw chicken sandwich was 10% instead of 8.4%?
4. What is the expected number of sandwiches to be bought before getting the first raw chicken sandwich?
5. How would you calculate the probability of getting exactly 2 raw chicken sandwiches in the first 5 buys?

**Tip:** Geometric distributions are useful for "waiting time" problems, where you're interested in how long it takes for a success to occur.

The probability of obtaining a raw chicken sandwich at Windies restaurant is 8.4%. Complete the table below to find the probability that the first raw chicken sandwich you get will be on the xth sandwich you buy.

Find out the probability of getting your first raw chicken sandwich on a specific trial using geometric distribution. This problem deals with an 8.4% chance of success and explores probability concepts with step-by-step calculations, perfect for Grades 10-12.

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