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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 Smith Ayala. in the United States
Jan 22,2025

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Answer

**c.** The probability of having at least one girl is \(1 - \frac{\binom{10}{4}}{\binom{22}{4}}\), which is approximately **97.13%**.

Solution

To find the probability that there will be **at least 1 girl** on the committee, it's often easier to use the complementary probability. Specifically, we'll calculate the probability that **all 4 members are boys** and then subtract this from 1. ### Step-by-Step Solution: 1. **Total Number of Students:** - Boys: 10 - Girls: 12 - **Total:** 10 + 12 = 22 students 2. **Total Number of Ways to Form a Committee of 4:** \[ \text{Total ways} = \binom{22}{4} \] 3. **Number of Ways to Form a Committee of 4 All Boys:** \[ \text{All boys} = \binom{10}{4} \] 4. **Probability All 4 Members Are Boys:** \[ P(\text{All boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} \] 5. **Probability of At Least 1 Girl:** \[ P(\text{At least 1 girl}) = 1 - P(\text{All boys}) = 1 - \frac{\binom{10}{4}}{\binom{22}{4}} \] ### Final Answer: **c.** The probability of having at least one girl is \[ 1 - \frac{\dbinom{10}{4}}{\dbinom{22}{4}} \] Numerically, this can be calculated as: \[ 1 - \frac{210}{7315} \approx 1 - 0.0287 = 0.9713 \] So, the probability is approximately **0.9713** or **97.13%**. **Answer:** Problem c Answer The probability is one minus (10 choose 4) divided by (22 choose 4). Thus, Probability = 1 − (10 C 4)/(22 C 4)

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First, let's break it down step by step! a. To find the probability that all 4 members on the committee will be girls, we need to calculate the number of ways to choose 4 girls from the 12 available. This can be done using combinations: \[ \text{Number of ways to choose 4 girls} = \binom{12}{4} = \frac{12!}{4!(12-4)!} = 495 \] Next, we calculate the total number of ways to choose any 4 students from the group of 22 students (10 boys + 12 girls): \[ \text{Total ways to choose 4 students} = \binom{22}{4} = \frac{22!}{4!(22-4)!} = 7315 \] So, the probability that all 4 committee members are girls is: \[ P(\text{all girls}) = \frac{\binom{12}{4}}{\binom{22}{4}} = \frac{495}{7315} \approx 0.0677 \] (Would be about 6.77%) --- b. To find the probability that all 4 members of the committee will be boys, we use a similar approach: \[ \text{Number of ways to choose 4 boys} = \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \] Using the previously calculated total number of ways to choose 4 students (which is still 7315), we find: \[ P(\text{all boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} = \frac{210}{7315} \approx 0.0287 \] (Would be about 2.87%) --- c. To calculate the probability that there is at least 1 girl in the committee, it's easier to use the complement rule. First, we can find the probability of having no girls (which would mean all boys): \[ P(\text{no girls}) = P(\text{all boys}) = \frac{\binom{10}{4}}{\binom{22}{4}} \approx 0.0287 \] Now we can find the probability of at least 1 girl: \[ P(\text{at least 1 girl}) = 1 - P(\text{no girls}) = 1 - 0.0287 \approx 0.9713 \] (Would be about 97.13%) --- In summary: - Probability all members are girls: ~6.77% - Probability all members are boys: ~2.87% - Probability at least 1 member is a girl: ~97.13%

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