solve: You have a bowl of 35 different mini candy bars. There are 16 Snickers, 11 Mars, and 8 Almond Joys. Without looking you go and grab some candy. 

No replacement:  What is the probability of getting 1 Snickers and 1 Almond Joy assuming that the first one is not replaced?  

 

Replacement:  What is the probability of getting 1 Snickers and 1 Almond Joy assuming that the first one is replaced?

What difference does it make?  Which gives the higher probability? 

Let's break down the two scenarios step by step.

### 1. **No Replacement**:

You have 35 candy bars, and you want to calculate the probability of getting 1 Snickers first and 1 Almond Joy second without replacement.

- **Step 1: Probability of drawing 1 Snickers first.**
  There are 16 Snickers out of 35 total candy bars, so the probability is:
  \[
  P(\text{Snickers first}) = \frac{16}{35}
  \]

- **Step 2: Probability of drawing 1 Almond Joy second (after the first one is taken out).**
  After removing one Snickers, you are left with 34 candy bars. There are still 8 Almond Joys left, so the probability is:
  \[
  P(\text{Almond Joy second}) = \frac{8}{34}
  \]

- **Total Probability (No Replacement)**: Multiply the two probabilities:
  \[
  P(\text{Snickers and Almond Joy}) = \frac{16}{35} \times \frac{8}{34} = \frac{128}{1190} \approx 0.1076
  \]

### 2. **With Replacement**:

In this case, after drawing the first candy (a Snickers), you put it back in the bowl, so you again have 35 candies in total for the second draw.

- **Step 1: Probability of drawing 1 Snickers first.**
  \[
  P(\text{Snickers first}) = \frac{16}{35}
  \]

- **Step 2: Probability of drawing 1 Almond Joy second (with replacement, so still 35 candies).**
  \[
  P(\text{Almond Joy second}) = \frac{8}{35}
  \]

- **Total Probability (With Replacement)**: Multiply the two probabilities:
  \[
  P(\text{Snickers and Almond Joy}) = \frac{16}{35} \times \frac{8}{35} = \frac{128}{1225} \approx 0.1045
  \]

### 3. **Difference and Comparison**:

- **Without Replacement**: The probability is approximately **0.1076** (10.76%).
- **With Replacement**: The probability is approximately **0.1045** (10.45%).

The probability is slightly higher **without replacement** because after removing one Snickers, the total number of candies decreases, making it a little more likely to pick an Almond Joy on the second draw.

### Conclusion:
The scenario **without replacement** gives a slightly higher probability of drawing 1 Snickers and 1 Almond Joy.

---

Would you like more details or have any further questions? Here are 5 related questions you might find interesting:

1. What is the probability of drawing 2 Almond Joys without replacement?
2. How does the probability change if the bowl has more than 35 candies but the same proportion of each type?
3. What if you wanted to draw 1 Mars and 1 Snickers without replacement?
4. How do these probabilities compare if you are allowed to look while picking the candies?
5. What is the probability of drawing 1 Almond Joy first and 1 Snickers second without replacement?

**Tip**: Always consider how "with replacement" or "without replacement" affects the total number of items when calculating probabilities.

You have a bowl of 35 different mini candy bars. There are 16 Snickers, 11 Mars, and 8 Almond Joys. Without looking, you go and grab some candy. No replacement: What is the probability of getting 1 Snickers and 1 Almond Joy assuming that the first one is not replaced? Replacement: What is the probability of getting 1 Snickers and 1 Almond Joy assuming that the first one is replaced? What difference does it make? Which gives the higher probability?

Understand the probability of drawing 1 Snickers and 1 Almond Joy from a bowl of 35 candy bars (16 Snickers, 11 Mars, and 8 Almond Joys) both with and without replacement. This problem explores core concepts of conditional probability and combinatorics, providing a step-by-step solution for students in Grades 9-12.

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