From a group of 10 boys and 12 girls, a committee of 4 students is chosen at random.
a. What is the probability that all 4 members on the committee will be girls?
b. What is the probability that all 4 members of the committee will be boys?
c. What is the probability that there will be at least 1 girl on the committee?

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UpStudy AI · Accepted Answer

To determine the probability that all 4 members of the committee are girls, we'll use combinatorial analysis.

**a. Probability that all 4 members are girls**

1. **Total number of students:**  
   - Boys: 10  
   - Girls: 12  
   - **Total:** 10 + 12 = **22**

2. **Total number of ways to choose 4 students out of 22:**  
   $$
   \text{Total ways} = \binom{22}{4} = \frac{22!}{4! \times (22-4)!} = 7315
   $$

3. **Number of ways to choose 4 girls out of 12:**  
   $$
   \text{Ways to choose girls} = \binom{12}{4} = \frac{12!}{4! \times (12-4)!} = 495
   $$

4. **Probability that all 4 members are girls:**  
   $$
   P(\text{all girls}) = \frac{\text{Ways to choose girls}}{\text{Total ways}} = \frac{495}{7315}
   $$

5. **Simplifying the fraction:**  
   $$
   \frac{495}{7315} = \frac{99}{1463} = \frac{9}{133} \approx 0.0677 \text{ or } 6.77\%
   $$

**Final Answer:**

Problem a Answer:

The probability is $$\frac{9}{133}$$ or approximately 6.77 %.
The probability that all 4 committee members are girls is $$\frac{9}{133}$$ or about 6.77%. [Continuing to the next parts] **b. Probability that all 4 members are boys** 1. **Number of ways to choose 4 boys out of 10:** $$ \text{Ways to choose boys} = \binom{10}{4} = \frac{10!}{4! \times (10-4)!} = 210 $$ 2. **Probability that all 4 members are boys:** $$ P(\text{all boys}) = \frac{210}{7315} = \frac{42}{1463} \approx 0.0287 \text{ or } 2.87\% $$ **Final Answer:** The probability that all 4 committee members are boys is $$\frac{42}{1463}$$ or approximately 2.87%. **c. Probability that there is at least 1 girl on the committee** 1. **Probability that all 4 members are boys (from part b):** $$ P(\text{all boys}) = \frac{42}{1463} \approx 2.87\% $$ 2. **Probability that there is at least 1 girl:** $$ P(\text{at least 1 girl}) = 1 - P(\text{all boys}) = 1 - \frac{42}{1463} = \frac{1421}{1463} \approx 0.9713 \text{ or } 97.13\% $$ **Final Answer:** The probability that there is at least 1 girl on the committee is $$\frac{1421}{1463}$$ or approximately 97.13%.

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