We will calculate the probability of selecting a committee of 4 students that contains at least 3 girls from a class of 10 boys and 5 girls.

Step 1: Determine the total ways to select 4 students

The total number of students is: $10 + 5 = 15$ The number of ways to choose any 4 students from 15 is given by the combination formula: $\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!}$

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

Step 2: Find the favorable cases (at least 3 girls)

We consider two cases:

1. Case 1: Exactly 3 girls and 1 boy

   • Choose 3 girls from 5: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$
   • Choose 1 boy from 10: $\binom{10}{1} = 10$
   • Total ways for this case: $10 \times 10 = 100$

2. Case 2: Exactly 4 girls

   • Choose 4 girls from 5: $\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5$
   • Total ways for this case: $5$

Thus, the total number of favorable outcomes: $100 + 5 = 105$

Step 3: Compute the probability

$P(\text{at least 3 girls}) = \frac{\text{favorable cases}}{\text{total cases}} = \frac{105}{1365}$

Simplifying:

$P = \frac{105}{1365} = \frac{7}{91} \approx 0.077$

Thus, the probability of selecting a committee with at least 3 girls is 0.077 (or 7.7%).

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Would you like a more detailed explanation on any step?

Here are some related questions you might find interesting:

1. What is the probability of selecting a committee with at most 2 girls?
2. What is the probability of selecting exactly 2 boys and 2 girls?
3. What is the probability of selecting a committee with all boys?
4. How does the probability change if there were 6 girls instead of 5?
5. How many different committees can be formed with exactly 3 boys?

Tip:

When dealing with probability problems involving selections, always break them down into favorable outcomes and total outcomes using combinations.