Suppose that you have 9 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement.

G1 = the first card drawn is green
G2 = the second card drawn is green

a.  P(G1 and G2) =

b.  P(At least 1 green) =

c.  P(G2|G1) =

d.  Are G1 and G2 independent?

They are independent events
They are dependent events

Let's solve the problem step by step.

### Given:
- Total green cards = 9
- Total yellow cards = 5
- Total cards = 9 + 5 = 14

### a. P(G1 and G2)

To find \( P(G1 \text{ and } G2) \):

1. Probability that the first card is green (G1):
   \[
   P(G1) = \frac{9}{14}
   \]

2. If the first card is green, there are now 8 green cards left out of 13 total cards. Thus, the probability that the second card is green (G2) given the first was green:
   \[
   P(G2 | G1) = \frac{8}{13}
   \]

3. Therefore, using the multiplication rule:
   \[
   P(G1 \text{ and } G2) = P(G1) \times P(G2 | G1) = \frac{9}{14} \times \frac{8}{13} = \frac{72}{182} = \frac{36}{91}
   \]

### b. P(At least 1 green)

We can find \( P(\text{at least 1 green}) \) by calculating the complement of the probability of drawing no green cards (i.e., both cards are yellow).

1. Probability that the first card is yellow:
   \[
   P(\text{1st yellow}) = \frac{5}{14}
   \]

2. If the first card is yellow, the probability that the second card is yellow:
   \[
   P(\text{2nd yellow | 1st yellow}) = \frac{4}{13}
   \]

3. Thus, the probability of drawing two yellow cards:
   \[
   P(\text{both yellow}) = P(\text{1st yellow}) \times P(\text{2nd yellow | 1st yellow}) = \frac{5}{14} \times \frac{4}{13} = \frac{20}{182} = \frac{10}{91}
   \]

4. Therefore:
   \[
   P(\text{at least 1 green}) = 1 - P(\text{both yellow}) = 1 - \frac{10}{91} = \frac{81}{91}
   \]

### c. P(G2 | G1)

From part (a), we already determined that:
\[
P(G2 | G1) = \frac{8}{13}
\]

### d. Are G1 and G2 independent?

Two events \( G1 \) and \( G2 \) are independent if:
\[
P(G2 | G1) = P(G2)
\]

1. Calculate \( P(G2) \):
   - The total ways to choose a green card in the second draw (irrespective of the first draw):
   \[
   P(G2) = \frac{9}{14} \text{ (if first is yellow)} + \frac{8}{14} \cdot \frac{5}{13} \text{ (if first is green)}
   \]
   However, the calculation is complex here. More simply, since \( P(G2 | G1) \neq P(G2) \) (the probability changes based on whether the first card was green), we conclude that:

**G1 and G2 are dependent events.**

### Summary:
a. \( P(G1 \text{ and } G2) = \frac{36}{91} \)  
b. \( P(\text{At least 1 green}) = \frac{81}{91} \)  
c. \( P(G2 | G1) = \frac{8}{13} \)  
d. They are dependent events.

Would you like more details or have any questions? Here are some related questions to consider:

1. What would happen if we had more yellow cards?
2. How would the probabilities change if we drew three cards instead of two?
3. What is the probability of drawing two yellow cards?
4. How does the order of drawing cards affect the probabilities?
5. Can you explain the concept of independent events in another context? 

**Tip:** When dealing with probability problems involving conditional probabilities, always start by identifying the relevant events and whether they are dependent or independent.

Suppose that you have 9 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement. a. P(G1 and G2) = b. P(At least 1 green) = c. P(G2|G1) = d. Are G1 and G2 independent?

Explore the probability of drawing two cards from a deck of 9 green and 5 yellow cards. Learn how to calculate P(G1 and G2), P(at least 1 green), and whether G1 and G2 are independent events. Detailed solutions provided.

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