magine you have a bag containing 72 red gummy bears and 45 green gummy bears (117 gummy bears total).

(a) With replacement:
You want to find the probability of drawing two green gummy bears in a row with replacement. This means after drawing the first gummy bear, you put it back in the bag before drawing again.

(b) Without replacement:
Now, find the probability of drawing two green gummy bears in a row without replacement. This means after the first gummy bear is drawn, you do not put it back into the bag before drawing the second one.

Let's break down both parts of the problem involving probability with and without replacement.

### (a) With Replacement
When drawing **with replacement**, the total number of gummy bears stays the same (117), and the probability for each draw is independent of the previous one.

- The probability of drawing a green gummy bear on the first draw is:

\[
P(\text{first green}) = \frac{45}{117}
\]

- Since the gummy bear is replaced, the probability of drawing a green gummy bear on the second draw is also:

\[
P(\text{second green}) = \frac{45}{117}
\]

Now, to find the probability of drawing **two green gummy bears in a row**, we multiply these probabilities:

\[
P(\text{two green with replacement}) = \frac{45}{117} \times \frac{45}{117}
\]

\[
P(\text{two green with replacement}) = \left(\frac{45}{117}\right)^2 = \frac{2025}{13689} \approx 0.1479
\]

Thus, the probability of drawing two green gummy bears in a row **with replacement** is approximately **0.1479**, or about **14.79%**.

### (b) Without Replacement
When drawing **without replacement**, the total number of gummy bears decreases after each draw, and the probabilities change as a result.

- The probability of drawing a green gummy bear on the first draw is:

\[
P(\text{first green}) = \frac{45}{117}
\]

- After drawing a green gummy bear, there are now 44 green gummy bears left, and only 116 total gummy bears in the bag. So, the probability of drawing a green gummy bear on the second draw is:

\[
P(\text{second green}) = \frac{44}{116}
\]

Now, to find the probability of drawing **two green gummy bears in a row without replacement**, we multiply these probabilities:

\[
P(\text{two green without replacement}) = \frac{45}{117} \times \frac{44}{116}
\]

\[
P(\text{two green without replacement}) = \frac{45 \times 44}{117 \times 116} = \frac{1980}{13572} \approx 0.1459
\]

Thus, the probability of drawing two green gummy bears in a row **without replacement** is approximately **0.1459**, or about **14.59%**.

### Summary
- **With replacement**: The probability of drawing two green gummy bears is **0.1479** or **14.79%**.
- **Without replacement**: The probability of drawing two green gummy bears is **0.1459** or **14.59%**.

---

Would you like more details or have any questions?

Here are five related questions to explore:

1. What is the probability of drawing two red gummy bears in a row with replacement?
2. What is the probability of drawing one green and one red gummy bear in any order with replacement?
3. How does the probability change if the number of green gummy bears is increased?
4. What is the probability of drawing three green gummy bears in a row without replacement?
5. How would the probabilities change if the total number of gummy bears were halved?

**Tip**: When dealing with probabilities involving "without replacement," remember that the total number of items and their composition changes after each draw.

Imagine you have a bag containing 72 red gummy bears and 45 green gummy bears (117 gummy bears total).

(a) With replacement:
You want to find the probability of drawing two green gummy bears in a row with replacement. This means after drawing the first gummy bear, you put it back in the bag before drawing again.

(b) Without replacement:
Now, find the probability of drawing two green gummy bears in a row without replacement. This means after the first gummy bear is drawn, you do not put it back into the bag before drawing the second one.

Find the probability of drawing two green gummy bears in a row from a bag containing 117 gummy bears (72 red, 45 green), with and without replacement. Detailed step-by-step solution provided using probability rules. Suitable for Grades 9-12.

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