To find the probability of selecting a committee of 4 females from a group consisting of 19 men and 11 women, we can use combinatorics.

Step-by-Step Solution

1. Total population:

   • Men: 19
   • Women: 11
   • Total: 19 + 11 = 30 people

2. Choosing 4 females from 11 women: This is a combination problem, and the formula for combinations is: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ where $n$ is the total number of elements and $r$ is the number of elements chosen.

   So, the number of ways to choose 4 women from 11 women is: $\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$

3. Choosing the rest (0 men from 19): Since we are only selecting women and no men, we need to calculate the number of ways to choose 0 men from 19 men: $\binom{19}{0} = 1$

4. Total number of ways to select any 4 people from 30 people: $\binom{30}{4} = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,405$

5. Probability: The probability is the ratio of favorable outcomes (selecting 4 women) to the total outcomes (selecting any 4 people): $P(\text{4 women}) = \frac{\binom{11}{4} \times \binom{19}{0}}{\binom{30}{4}} = \frac{330 \times 1}{27,405} = \frac{330}{27,405} \approx 0.01204$

Thus, the probability of selecting a committee of 4 females is approximately 0.01204 or 1.204%.

---

Would you like more details or have any questions about this problem?

Here are some related questions for further exploration:

1. What is the probability of selecting 3 females and 1 male?
2. How does the probability change if there are 15 men and 15 women?
3. What is the probability of selecting at least 1 female?
4. How would the result change if we select a committee of 5 instead of 4?
5. How many different committees of 4 people can be formed from the group?

Tip: In probability problems involving selection, combinations are often used to find the total possible outcomes without worrying about the order.