To solve the problem, let's denote the total number of cards as $n = 4 + 5 = 9$.

a. $P(G1 \text{ and } G2)$

To find the probability that both drawn cards are green:

1. Probability that the first card is green (G1): $P(G1) = \frac{4}{9}$

2. Probability that the second card is green (G2) given that the first was green: After drawing one green card, there are 3 green cards left and 8 cards in total: $P(G2 | G1) = \frac{3}{8}$

Now, using the multiplication rule for probabilities: $P(G1 \text{ and } G2) = P(G1) \times P(G2 | G1) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}$

b. $P(\text{At least 1 green})$

To find the probability of drawing at least one green card, it’s easier to calculate the complement, which is the probability of drawing no green cards (i.e., both cards are yellow):

1. Probability that the first card is yellow: $P(\text{1st yellow}) = \frac{5}{9}$

2. Probability that the second card is yellow given the first was yellow: After drawing one yellow card, there are 4 yellow cards left and 8 cards total: $P(\text{2nd yellow | 1st yellow}) = \frac{4}{8} = \frac{1}{2}$

Now, calculate the probability of both cards being yellow: $P(\text{Both yellow}) = P(\text{1st yellow}) \times P(\text{2nd yellow | 1st yellow}) = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}$

Thus, the probability of at least one green card is: $P(\text{At least 1 green}) = 1 - P(\text{Both yellow}) = 1 - \frac{5}{18} = \frac{13}{18}$

c. $P(G2 | G1)$

This was calculated earlier as part of $P(G1 \text{ and } G2)$: $P(G2 | G1) = \frac{3}{8}$

d. Are $G1$ and $G2$ independent?

Two events $A$ and $B$ are independent if: $P(A \text{ and } B) = P(A) \times P(B)$

• We have $P(G1 \text{ and } G2) = \frac{1}{6}$.

• Calculate $P(G2)$:

  • The probability that the second card is green can be found by considering all possible cases for the first card:
    1. If the first card is green: $\frac{4}{9} \times \frac{3}{8}$
    2. If the first card is yellow: $\frac{5}{9} \times \frac{4}{8}$

  Thus, $P(G2) = P(G2 | G1) P(G1) + P(G2 | \text{not } G1) P(\text{not } G1) = \left(\frac{3}{8} \cdot \frac{4}{9}\right) + \left(\frac{4}{8} \cdot \frac{5}{9}\right) = \frac{12}{72} + \frac{20}{72} = \frac{32}{72} = \frac{4}{9}$

Now we can check independence: $P(G1) \times P(G2) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$

Since $P(G1 \text{ and } G2) = \frac{1}{6}$ is not equal to $\frac{16}{81}$, $G1$ and $G2$ are dependent events.

Summary

• a. $P(G1 \text{ and } G2) = \frac{1}{6}$
• b. $P(\text{At least 1 green}) = \frac{13}{18}$
• c. $P(G2 | G1) = \frac{3}{8}$
• d. They are dependent events.

Would you like further details on any part of this problem? Here are some follow-up questions:

1. What would be the probability if there were more cards?
2. How does drawing with replacement change these probabilities?
3. What other combinations of card draws could we analyze?
4. How would the probabilities change if the number of yellow cards increased?
5. Can you explain the concept of independent events further?

Tip: Always consider the total number of outcomes when dealing with probabilities to ensure accuracy.