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Part 1
From a group of 11 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?
Question content area bottom
Part 1
a. The probability that all 5 members on the committee will be girls is 
  
enter your response here.
​(Type an integer or a simplified​ fraction.)
Part 2
b. The probability that all 5 members on the committee will be boys is 
  
enter your response here.
​(Type an integer or a simplified​ fraction.)
Part 3
c. The probability that there will be at least 1 girl on the committee is 
  
enter your response here.
​(Type an integer or a simplified​ fraction.)

We are asked to calculate the probabilities related to selecting a committee of 5 students from a group of 11 boys and 13 girls.

### Total number of students:
\[
11 \text{ boys} + 13 \text{ girls} = 24 \text{ students}
\]
### Total possible ways to form a committee of 5 students from 24:
The number of ways to choose 5 students from 24 is calculated using combinations:
\[
\binom{24}{5} = \frac{24!}{5!(24-5)!} = \frac{24!}{5!19!}
\]

Now let's go step-by-step through each part of the question.

### Part 1 (a): Probability that all 5 members on the committee are girls

The number of ways to choose 5 girls from the 13 available girls:
\[
\binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13!}{5!8!}
\]

The probability that all 5 members are girls is:
\[
\text{Probability} = \frac{\binom{13}{5}}{\binom{24}{5}}
\]

### Part 2 (b): Probability that all 5 members on the committee are boys

The number of ways to choose 5 boys from the 11 available boys:
\[
\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!}
\]

The probability that all 5 members are boys is:
\[
\text{Probability} = \frac{\binom{11}{5}}{\binom{24}{5}}
\]

### Part 3 (c): Probability that there will be at least 1 girl on the committee

The complement of this event is that there are no girls, meaning all 5 members are boys. We already computed the probability that all 5 members are boys in part (b), so the probability of at least 1 girl is:
\[
\text{Probability} = 1 - \text{(Probability that all 5 members are boys)}
\]

Let me calculate these values.### Part 1 (a):
The probability that all 5 members of the committee will be girls is approximately:
\[
0.0303
\]

### Part 2 (b):
The probability that all 5 members of the committee will be boys is approximately:
\[
0.0109
\]

### Part 3 (c):
The probability that there will be at least 1 girl on the committee is approximately:
\[
0.9891
\]

Would you like more details on how these probabilities were calculated or have any specific questions?

Here are 5 related questions to deepen your understanding:
1. What is the probability of forming a committee with exactly 3 girls and 2 boys?
2. How does the probability change if the group has more boys than girls?
3. Can you explain how the complement rule helps in calculating the probability of at least 1 girl?
4. How would the probabilities change if the committee had 6 members instead of 5?
5. What is the probability of selecting a committee with at least 2 boys?

**Tip**: Always double-check if the question asks for a simplified fraction or a decimal approximation when solving probability problems.

From a group of 11 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?

Calculate the probability of selecting a committee of 5 students from 11 boys and 13 girls, including finding the probability of choosing all girls, all boys, and at least one girl. This detailed step-by-step guide covers key probability and combinatorics concepts. Ideal for students in Grades 9-12.

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