Question 16: a) Define P(A ∩ B) for independent events. b) Two cards are drawn without replacement. Find the probability both are face cards. c) Draw a tree diagram for face card probabilities.

Here's the solution to question 16:

a) If two events $A$ and $B$ are independent events, the formula for finding $P(A \cap B)$ is:

$$P(A \cap B) = P(A) \times P(B)$$

The final answer is $\boxed{P(A \cap B) = P(A) \times P(B)}$.

b) Two cards are drawn randomly one after another without replacement.
Total number of cards = 52.
Number of face cards (King, Queen, Jack) = 3 cards/suit $\times$ 4 suits = 12 face cards.

Step 1: Calculate the probability of the first card being a face card.

$$P(\text{1st Face}) = \frac{\text{Number of face cards}}{\text{Total cards}} = \frac{12}{52} = \frac{3}{13}$$

Step 2: Calculate the probability of the second card being a face card, given the first was a face card and not replaced.
After drawing one face card, there are 51 cards remaining, and 11 face cards remaining.

$$P(\text{2nd Face | 1st Face}) = \frac{\text{Remaining face cards}}{\text{Remaining cards}} = \frac{11}{51}$$

Step 3: Calculate the probability of both cards being face cards.

$$P(\text{Both Face}) = P(\text{1st Face}) \times P(\text{2nd Face | 1st Face})$$ $$P(\text{Both Face}) = \frac{12}{52} \times \frac{11}{51} = \frac{3}{13} \times \frac{11}{51} = \frac{33}{663} = \frac{11}{221}$$

The final answer is $\boxed{\frac{11}{221}}$.

c) Show the probability of all the possible outcomes of getting or not-getting a face card in a tree diagram (without replacement).

Let $F$ be the event of getting a face card.
Let $F'$ be the event of not getting a face card.
Number of face cards = 12.
Number of non-face cards = $52 - 12 = 40$.

Tree Diagram:

First Draw:

• Probability of getting a face card ($F_1$): $P(F_1) = \frac{12}{52} = \frac{3}{13}$
• Probability of not getting a face card ($F'_1$): $P(F'_1) = \frac{40}{52} = \frac{10}{13}$

Second Draw (without replacement):

• If the first card was a face card ($F_1$):

  • Remaining cards = 51
  • Remaining face cards = 11
  • Remaining non-face cards = 40
  • Probability of getting a face card ($F_2 | F_1$): $P(F_2 | F_1) = \frac{11}{51}$
  • Probability of not getting a face card ($F'_2 | F_1$): $P(F'_2 | F_1) = \frac{40}{51}$

• If the first card was not a face card ($F'_1$):

  • Remaining cards = 51
  • Remaining face cards = 12
  • Remaining non-face cards = 39
  • Probability of getting a face card ($F_2 | F'_1$): $P(F_2 | F'_1) = \frac{12}{51} = \frac{4}{17}$
  • Probability of not getting a face card ($F'_2 | F'_1$): $P(F'_2 | F'_1) = \frac{39}{51} = \frac{13}{17}$

Possible Outcomes and their Probabilities:

1. Both Face Cards ($F_1$ and $F_2$):

   $$P(F_1 \cap F_2) = P(F_1) \times P(F_2 | F_1) = \frac{12}{52} \times \frac{11}{51} = \frac{3}{13} \times \frac{11}{51} = \frac{11}{221}$$

2. First Face, Second Not Face ($F_1$ and $F'_2$):

   $$P(F_1 \cap F'_2) = P(F_1) \times P(F'_2 | F_1) = \frac{12}{52} \times \frac{40}{51} = \frac{3}{13} \times \frac{40}{51} = \frac{40}{221}$$

3. First Not Face, Second Face ($F'_1$ and $F_2$):

   $$P(F'_1 \cap F_2) = P(F'_1) \times P(F_2 | F'_1) = \frac{40}{52} \times \frac{12}{51} = \frac{10}{13} \times \frac{12}{51} = \frac{40}{221}$$

4. Both Not Face Cards ($F'_1$ and $F'_2$):

   $$P(F'_1 \cap F'_2) = P(F'_1) \times P(F'_2 | F'_1) = \frac{40}{52} \times \frac{39}{51} = \frac{10}{13} \times \frac{39}{51} = \frac{130}{221}$$

The tree diagram visually represents these branches and their associated probabilities.

d) If two cards are drawn randomly one after another with replacement.

Step 1: Calculate the probability that both cards are face cards (with replacement).
Number of face cards = 12.

$$P(\text{Both Face, with replacement}) = P(\text{1st Face}) \times P(\text{2nd Face})$$ $$P(\text{Both Face, with replacement}) = \frac{12}{52} \times \frac{12}{52} = \frac{3}{13} \times \frac{3}{13} = \frac{9}{169}$$

Step 2: Calculate the probability that both cards are ace (with replacement).
Number of ace cards = 4.

$$P(\text{Both Ace, with replacement}) = P(\text{1st Ace}) \times P(\text{2nd Ace})$$ $$P(\text{Both Ace, with replacement}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$$

Step 3: Determine how many times more is the probability that both are face cards than the probability that both cards are ace.

$$$$