To solve this problem, we need to calculate the probability that a committee of 4 students selected from a class of 10 boys and 5 girls contains at least 3 girls.

Step 1: Total number of ways to select 4 students

The total number of ways to select 4 students from the 15 students (10 boys + 5 girls) is given by the combination formula:

$\text{Total ways} = \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$

Step 2: Favorable outcomes (At least 3 girls)

The committee can have:

• 3 girls and 1 boy, or
• 4 girls.

We will calculate the number of ways for each scenario.

Case 1: 3 girls and 1 boy

The number of ways to choose 3 girls from 5 is:

$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$

The number of ways to choose 1 boy from 10 is:

$\binom{10}{1} = 10$

So, the total number of ways to select 3 girls and 1 boy is:

$\binom{5}{3} \times \binom{10}{1} = 10 \times 10 = 100$

Case 2: 4 girls

The number of ways to choose 4 girls from 5 is:

$\binom{5}{4} = \frac{5}{1} = 5$

Step 3: Total number of favorable outcomes

The total number of favorable outcomes is the sum of the two cases:

$\text{Favorable outcomes} = 100 + 5 = 105$

Step 4: Probability calculation

The probability is the ratio of favorable outcomes to total outcomes:

$P(\text{At least 3 girls}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{105}{1365}$

Simplifying this fraction:

$P(\text{At least 3 girls}) = \frac{7}{91} \approx 0.0769$

So, the probability that the committee contains at least 3 girls is approximately 0.0769 or 7.69%.

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Would you like more details on the steps or another example? Here are some related questions for further exploration:

1. How would the probability change if there were 6 girls and 9 boys?
2. What is the probability that a committee of 4 contains no girls?
3. If a committee of 5 is selected, what is the probability that it contains at least 3 girls?
4. How do we calculate the probability for the committee to contain exactly 2 boys and 2 girls?
5. How do you generalize this probability calculation for committees of any size from any group of students?

Tip: Remember that the combination formula $\binom{n}{k}$ is used to find the number of ways to choose $k$ objects from $n$ objects without regard to order, and it can be calculated as $\frac{n!}{k!(n-k)!}$.