There are 9 apples, 6 oranges, 3 pears, and 4 peaches in Grocery Bag.

What is the probability of drawing two peaches from the bag assuming the first one is replaced?

What is the probability of two peaches from the bag assuming the first one has been replaced?

What difference does it make?

Which gives the higher probability?

Let's break down the problem and solve for each scenario:

### Total Number of Fruits
There are 9 apples, 6 oranges, 3 pears, and 4 peaches in the grocery bag. The total number of fruits is:

\[
9 + 6 + 3 + 4 = 22 \text{ fruits}
\]

### Probability of Drawing Two Peaches (With Replacement)
When the first peach is replaced, the total number of fruits remains the same for both draws (22 fruits). The probability of drawing a peach on each draw is:

\[
P(\text{Peach on 1st draw}) = \frac{4}{22}
\]
Since the first fruit is replaced, the second draw has the same probability:
\[
P(\text{Peach on 2nd draw}) = \frac{4}{22}
\]

The probability of drawing two peaches with replacement is:
\[
P(\text{Two Peaches with Replacement}) = \frac{4}{22} \times \frac{4}{22} = \frac{16}{484} = \frac{4}{121} \approx 0.033
\]

### Probability of Drawing Two Peaches (Without Replacement)
In this scenario, after drawing the first peach, it is **not** replaced. Therefore, the total number of fruits decreases by 1 (from 22 to 21), and the number of peaches decreases by 1 (from 4 to 3). The probabilities for the two draws are as follows:

\[
P(\text{Peach on 1st draw}) = \frac{4}{22}
\]
\[
P(\text{Peach on 2nd draw}) = \frac{3}{21}
\]

The probability of drawing two peaches without replacement is:
\[
P(\text{Two Peaches without Replacement}) = \frac{4}{22} \times \frac{3}{21} = \frac{12}{462} = \frac{2}{77} \approx 0.026
\]

### Comparison: Which Gives the Higher Probability?
- **With Replacement**: \( \frac{4}{121} \approx 0.033 \)
- **Without Replacement**: \( \frac{2}{77} \approx 0.026 \)

The probability of drawing two peaches is **higher when the first fruit is replaced**.

### Why is there a difference?
When the first peach is replaced, the probabilities remain the same for both draws, making it more likely to draw two peaches. Without replacement, the number of peaches and the total number of fruits decrease after the first draw, lowering the chance of drawing a second peach.

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Would you like further details on these calculations, or have any questions?

Here are 5 related questions to expand on this:

1. What is the probability of drawing two apples without replacement?
2. How would the probabilities change if there were 5 peaches instead of 4?
3. What is the probability of drawing one apple and one peach with replacement?
4. How does the probability change if the total number of fruits increases?
5. What is the probability of drawing two peaches if the bag contains only peaches?

**Tip:** When dealing with probability problems, be sure to carefully note whether events are independent (like with replacement) or dependent (like without replacement).

There are 9 apples, 6 oranges, 3 pears, and 4 peaches in a Grocery Bag. What is the probability of drawing two peaches from the bag assuming the first one is replaced? What is the probability of two peaches from the bag assuming the first one has been replaced? What difference does it make? Which gives the higher probability?

This problem explores the probability of drawing two peaches from a bag containing 9 apples, 6 oranges, 3 pears, and 4 peaches, with and without replacement. Learn the key concepts of independent and dependent events, and how the probability changes based on the scenario.

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