9. A committee of 12 is to be selected from 18 soccer players and 15 basketball players. What is the probability that the number of soccer players on the committee is triple the number of basketball players.



Let's solve the problem step by step.

### Problem Restatement:
We have 18 soccer players and 15 basketball players. A committee of 12 members is to be selected. We need to find the probability that the number of soccer players on the committee is exactly three times the number of basketball players.

### Step 1: Set Up the Equation

Let:
- \( s \) be the number of soccer players selected.
- \( b \) be the number of basketball players selected.

According to the problem:
\[ s + b = 12 \]
\[ s = 3b \]

### Step 2: Solve the System of Equations

Substitute \( s = 3b \) into \( s + b = 12 \):
\[ 3b + b = 12 \]
\[ 4b = 12 \]
\[ b = 3 \]

Since \( s = 3b \), then:
\[ s = 3 \times 3 = 9 \]

So, the committee should consist of 9 soccer players and 3 basketball players.

### Step 3: Calculate the Number of Favorable Outcomes

We need to calculate the number of ways to select 9 soccer players out of 18 and 3 basketball players out of 15.

The number of ways to select 9 soccer players out of 18 is:
\[ \binom{18}{9} = \frac{18!}{9!(18-9)!} \]

The number of ways to select 3 basketball players out of 15 is:
\[ \binom{15}{3} = \frac{15!}{3!(15-3)!} \]

The total number of favorable outcomes is:
\[ \binom{18}{9} \times \binom{15}{3} \]

### Step 4: Calculate the Total Number of Outcomes

The total number of ways to select any 12 players out of 33 (18 soccer + 15 basketball) is:
\[ \binom{33}{12} = \frac{33!}{12!(33-12)!} \]

### Step 5: Calculate the Probability

The probability that the committee has 9 soccer players and 3 basketball players is:
\[ P = \frac{\binom{18}{9} \times \binom{15}{3}}{\binom{33}{12}} \]

### Step 6: Compute the Values

Let's calculate the binomial coefficients and the probability.

- \(\binom{18}{9}\) = 48620
- \(\binom{15}{3}\) = 455
- \(\binom{33}{12}\) = 193536720

So,
\[ P = \frac{48620 \times 455}{193536720} \approx 0.114 \]

### Final Answer:
The probability that the committee has 9 soccer players and 3 basketball players is approximately \( 0.114 \), or 11.4%.

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Would you like further details on any of the steps, or do you have any other questions?

### Related Questions:
1. What is the probability of selecting exactly 6 soccer players and 6 basketball players?
2. How does the probability change if the committee size is increased to 15 members?
3. What if the selection is from only soccer players; what is the probability of choosing 12 soccer players?
4. How would the probability change if there are 20 soccer players instead of 18?
5. What if the committee requires at least one basketball player, how does that impact the probability?
6. How can we generalize the problem for different ratios of soccer to basketball players?
7. What is the total number of possible committees that can be formed from these 33 players?
8. How would the solution change if the basketball players are increased to 18?

**Tip:** When calculating probabilities involving combinations, simplifying the equations or steps before multiplying can help reduce computational errors.

Calculate the probability of forming a committee with exactly 9 soccer players and 3 basketball players from a total of 33 players (18 soccer players and 15 basketball players). Learn the step-by-step solution involving combinatorics and probability calculations.

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