Suppose that you have 5 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement. G_{1} = the first card drawn is green G_{2} = the second card drawn is green a. P(G_{1}) and (2) = b. P(At least 1 green) c. P(G2/G1) d. Are G_{1} and G_{2} independent? They are independent events They are dependent events

We are given a scenario with 5 green cards and 5 yellow cards, making a total of 10 cards. Let's address each part of the problem systematically:

### a. \( P(G_1) \)
\( P(G_1) \) represents the probability that the first card drawn is green.

Since there are 5 green cards out of a total of 10 cards:

\[
P(G_1) = \frac{5}{10} = \frac{1}{2}
\]

### b. \( P(\text{At least 1 green}) \)
This is the probability that at least one of the two cards drawn is green. To find this, we can use the complement rule: first calculate the probability that **neither card is green**, and then subtract this from 1.

- The probability that the first card is yellow is \( \frac{5}{10} = \frac{1}{2} \).
- If the first card is yellow, there are 4 yellow cards left, and 9 total cards remaining. Thus, the probability that the second card is also yellow is \( \frac{4}{9} \).

So, the probability that **both cards are yellow** (i.e., no green cards are drawn) is:

\[
P(\text{no green}) = \frac{5}{10} \times \frac{4}{9} = \frac{1}{2} \times \frac{4}{9} = \frac{2}{9}
\]

Therefore, the probability of drawing **at least 1 green card** is:

\[
P(\text{at least 1 green}) = 1 - P(\text{no green}) = 1 - \frac{2}{9} = \frac{7}{9}
\]

### c. \( P(G_2 | G_1) \)
\( P(G_2 | G_1) \) represents the conditional probability that the second card drawn is green, given that the first card was green.

- After drawing a green card first, there are now 4 green cards left out of a total of 9 cards.
  
Thus, the probability of drawing a second green card, given that the first was green, is:

\[
P(G_2 | G_1) = \frac{4}{9}
\]

### d. Are \( G_1 \) and \( G_2 \) independent events?

Two events are independent if the occurrence of one event does not affect the probability of the other. In this case, the probability of drawing a green card second (i.e., \( G_2 \)) depends on whether the first card drawn was green (i.e., \( G_1 \)).

Since \( P(G_2 | G_1) = \frac{4}{9} \), but \( P(G_2) \) without knowing the outcome of \( G_1 \) would simply be \( \frac{5}{10} = \frac{1}{2} \), the two probabilities are not the same. This means that \( G_1 \) and \( G_2 \) are **dependent events**.

### Conclusion:
- a. \( P(G_1) = \frac{1}{2} \)
- b. \( P(\text{at least 1 green}) = \frac{7}{9} \)
- c. \( P(G_2 | G_1) = \frac{4}{9} \)
- d. \( G_1 \) and \( G_2 \) are **dependent events**.

---

Would you like more details on any part of the explanation? Here are some related questions for further exploration:

1. What would \( P(\text{both green}) \) be?
2. How would the probabilities change if there were more yellow cards than green cards?
3. What is the general formula for conditional probability in this context?
4. How does replacement affect independence of events in probability?
5. What is the difference between dependent and independent events?

**Tip**: In conditional probability problems, it’s often helpful to calculate the total probability and then the conditional parts to better understand the relationships between events.

Explore the probability of drawing green and yellow cards, understanding concepts like conditional probability and dependent events. This problem is perfect for students in Grades 9-12, covering key theorems and formulas to solve probability problems without replacement.

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