Question

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?

Ask by Logan Davison. in the United States
Jan 22,2025

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Answer

The probability of having at least one girl on the committee is about 97.2%.

Solution

To determine the probability that there will be **at least 1 girl** on the committee, it's often easier to use the complement rule. The complement of having at least one girl is having **no girls at all** (i.e., all members are boys). **Step-by-Step Calculation:** 1. **Total number of students:** There are 10 boys and 12 girls, making a total of 22 students. 2. **Total number of ways to choose 4 students from 22:** \[ \binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315 \] 3. **Number of ways to choose 4 boys from 10:** \[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] 4. **Probability of choosing all boys (no girls):** \[ P(\text{All Boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} = \frac{210}{7315} \approx 0.0287 \text{ or } 2.87\% \] 5. **Probability of having at least 1 girl:** \[ P(\text{At least 1 Girl}) = 1 - P(\text{All Boys}) = 1 - 0.0287 = 0.9713 \text{ or } 97.13\% \] **Final Answer:** **Problem c Answer** The probability of having at least one girl on the committee is approximately 97.2 %.

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To answer your questions, let's calculate the probabilities step by step. a. First, find the total number of ways to choose 4 students from 22 (10 boys + 12 girls): \[ \text{Total ways} = \binom{22}{4} = \frac{22!}{4!(22-4)!} = 7315 \] Next, calculate the number of ways to choose 4 girls from 12: \[ \text{Ways to choose 4 girls} = \binom{12}{4} = \frac{12!}{4!(12-4)!} = 495 \] Now, the probability that all 4 members on the committee will be girls: \[ P(\text{all girls}) = \frac{\binom{12}{4}}{\binom{22}{4}} = \frac{495}{7315} \approx 0.0677 \] b. Similarly, calculate the number of ways to choose 4 boys from 10: \[ \text{Ways to choose 4 boys} = \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \] Now, the probability that all 4 members of the committee will be boys: \[ P(\text{all boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} = \frac{210}{7315} \approx 0.0287 \] c. To find the probability that there will be at least 1 girl, we can use the complement rule. The complement of "at least 1 girl" is "no girls", which means all boys in this case: \[ P(\text{at least 1 girl}) = 1 - P(\text{all boys}) \] Substituting the probability of all boys: \[ P(\text{at least 1 girl}) = 1 - \frac{210}{7315} = \frac{7315 - 210}{7315} = \frac{7105}{7315} \approx 0.9713 \] So, summarizing: - a. Probability that all members are girls: \( \approx 0.0677 \) - b. Probability that all members are boys: \( \approx 0.0287 \) - c. Probability that there is at least 1 girl: \( \approx 0.9713 \)

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