From a group of 10 boys and 13 girls, a committee of 5 students is chosen at random.
a. What is the probability that all 5 members on the committee will be girls?
b. What is the probability that all 5 members of the committee will be boys?
c. What is the probability that there will be at least 1 girl on the committee?
a. The probability that all 5 members on the committee will be girls is .
(Type an integer or a simplified fraction.)
b. The probability that all 5 members on the committee will be boys is .
(Type an integer or a simplified fraction.)
c. The probability that there will be at least 1 girl on the committee is .
(Type an integer or a simplified fraction.)

Ask by Bartlett Conner. in the United States

Jan 23,2025

Question

From a group of 10 boys and 13 girls, a committee of 5 students is chosen at random.
a. What is the probability that all 5 members on the committee will be girls?
b. What is the probability that all 5 members of the committee will be boys?
c. What is the probability that there will be at least 1 girl on the committee?
a. The probability that all 5 members on the committee will be girls is .
(Type an integer or a simplified fraction.)
b. The probability that all 5 members on the committee will be boys is .
(Type an integer or a simplified fraction.)
c. The probability that there will be at least 1 girl on the committee is .
(Type an integer or a simplified fraction.)

Ask by Bartlett Conner. in the United States

Jan 23,2025

UpStudy AI · Accepted Answer

To determine the probability that there will be **at least one girl** on the committee, we can use the complementary probability approach. This involves calculating the probability of the opposite event (i.e., **all committee members are boys**) and then subtracting that from 1.

### Step-by-Step Solution:

1. **Total Number of Students:**
   - Boys: 10
   - Girls: 13
   - **Total:** $$ 10 + 13 = 23 $$ students

2. **Total Number of Ways to Choose 5 Students:**
   $$
   \binom{23}{5} = \frac{23!}{5!(23-5)!} = 33,649
   $$

3. **Number of Ways to Choose 5 Boys:**
   $$
   \binom{10}{5} = \frac{10!}{5!(10-5)!} = 252
   $$
   - **Probability that all 5 members are boys:**
     $$
     P(\text{all boys}) = \frac{252}{33,649} = \frac{36}{4,807}
     $$

4. **Probability of At Least One Girl:**
   $$
   P(\text{at least one girl}) = 1 - P(\text{all boys}) = 1 - \frac{36}{4,807} = \frac{4,807 - 36}{4,807} = \frac{4,771}{4,807}
   $$
   - This fraction is already in its simplest form since 4,771 and 4,807 are coprime (their greatest common divisor is 1).

### Final Answer:
c. The probability that there will be at least 1 girl on the committee is $$ \frac{4771}{4807} $$.

**Answer:**

Problem c Answer:

The probability is $$ \frac{4771}{4807} $$.
The probability that there will be at least 1 girl on the committee is $$ \frac{4771}{4807} $$.

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