Suppose five cards are drawn from a standard 52 card deck of playing cards, as described in
Example 1.4.3, with all possibilities being equally likely.
(a) FLUSH is the event that all five cards have the same suit. Find P(FLUSH).
(b) SPECIAL is the event that the five cards are one of A, K, Q, J but not necessarily of
the same suit. Find P(SPECIAL).

Let's solve the two parts of the problem step-by-step.

### (a) Finding \( P(\text{FLUSH}) \)

A **flush** is a hand where all five cards have the same suit. In a standard deck of 52 cards, there are 4 suits (hearts, diamonds, clubs, and spades), each containing 13 cards.

1. **Number of ways to choose a flush**:  
   To form a flush, we first choose the suit (4 choices), and then we choose 5 cards from the 13 cards available in that suit.

   \[
   \text{Number of ways to choose 5 cards from 13 of the same suit} = \binom{13}{5}
   \]

2. **Total number of possible hands**:  
   The total number of ways to choose any 5 cards from the 52-card deck is:

   \[
   \binom{52}{5}
   \]

3. **Calculating the probability**:  
   The probability of drawing a flush is the ratio of the number of ways to draw a flush to the total number of possible hands.

   \[
   P(\text{FLUSH}) = \frac{\text{Number of flush hands}}{\text{Total number of hands}}
   = \frac{4 \cdot \binom{13}{5}}{\binom{52}{5}}
   \]

Now, let's compute this.

\[
\binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287
\]

\[
\binom{52}{5} = \frac{52!}{5!(52-5)!} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2598960
\]

Substitute these values:

\[
P(\text{FLUSH}) = \frac{4 \times 1287}{2598960} = \frac{5148}{2598960}
\]

\[
P(\text{FLUSH}) \approx 0.00198
\]

So, the probability of getting a flush is approximately **0.198%**.

### (b) Finding \( P(\text{SPECIAL}) \)

The **SPECIAL** event is defined as a hand containing any five cards that are among the ranks Ace (A), King (K), Queen (Q), and Jack (J), regardless of suits.

1. **Number of ways to form a SPECIAL hand**:  
   - We first identify the cards involved: A, K, Q, J. There are 4 ranks and 4 suits, meaning there are a total of \(4 \times 4 = 16\) cards to choose from (4 Aces, 4 Kings, 4 Queens, 4 Jacks).
   - We need to select 5 cards from these 16.

   \[
   \text{Number of SPECIAL hands} = \binom{16}{5}
   \]

2. **Total number of possible hands**:  
   As calculated previously:

   \[
   \binom{52}{5} = 2598960
   \]

3. **Calculating the probability**:  
   The probability of drawing a SPECIAL hand is:

   \[
   P(\text{SPECIAL}) = \frac{\binom{16}{5}}{\binom{52}{5}}
   \]

Now, compute \( \binom{16}{5} \):

\[
\binom{16}{5} = \frac{16!}{5!(16-5)!} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = 4368
\]

Substitute these values:

\[
P(\text{SPECIAL}) = \frac{4368}{2598960}
\]

\[
P(\text{SPECIAL}) \approx 0.00168
\]

So, the probability of getting a SPECIAL hand is approximately **0.168%**.

### Summary of Results

- **(a) \(P(\text{FLUSH})\)**: 0.198%
- **(b) \(P(\text{SPECIAL})\)**: 0.168%

Would you like more details or have any questions?  

Here are 5 related questions that could expand your understanding:
1. What is the probability of getting a straight in a 5-card hand?
2. How would the probability of getting a flush change if drawing 7 cards instead of 5?
3. What is the probability of getting a full house in a 5-card hand?
4. How does the calculation change if the deck has jokers included?
5. What is the probability of getting four of a kind in a 5-card hand?

**Tip**: When solving probability problems involving cards, always consider the total possible outcomes and how many of those outcomes satisfy the event condition.

Explore the probabilities of drawing flush and special hands from a standard 52-card deck. Understand how to calculate these probabilities step-by-step, including combinatorial calculations and probability ratios. Suitable for advanced high school students and enthusiasts of probability theory.

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