There are 5 Colorado beetles and 4 ladybugs in a jar. As a result of the jar falling, four insects fell out. Find the probabilities of the following events: A - only Colorado beetles fell out; B - 3 Colorado beetles and 1 ladybug fell out; C - among the fallen insects there are exactly 2 ladybugs.

1. Problem Description

There are 5 Colorado beetles and 4 ladybugs in a jar. As a result of the jar falling, four insects fell out. Find the probabilities of the following events:

A - only Colorado beetles fell out;

B - 3 Colorado beetles and 1 ladybug fell out;

C - among the fallen insects there are exactly 2 ladybugs.

2. Solution Steps

First, we need to find the total number of ways to choose 4 insects out of the 9 insects (5 beetles and 4 ladybugs). This can be calculated using combinations:

$$C(n, k) = \frac{n!}{k!(n-k)!}$$

where $n$ is the total number of items and $k$ is the number of items to choose.

The total number of ways to choose 4 insects from 9 is:

$$C(9, 4) = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$

A - Only Colorado beetles fell out. This means we chose 4 beetles out of the 5 available. The number of ways to do this is:

$$C(5, 4) = \frac{5!}{4!1!} = \frac{5 \times 4 \times 3 \times 2}{4 \times 3 \times 2 \times 1} = 5$$

The probability of event A is:

$$P(A) = \frac{C(5, 4)}{C(9, 4)} = \frac{5}{126}$$

B - 3 Colorado beetles and 1 ladybug fell out. We need to choose 3 beetles out of 5 and 1 ladybug out of

4. The number of ways to do this is:

$$C(5, 3) \times C(4, 1) = \frac{5!}{3!2!} \times \frac{4!}{1!3!} = \frac{5 \times 4}{2 \times 1} \times 4 = 10 \times 4 = 40$$

The probability of event B is:

$$P(B) = \frac{C(5, 3) \times C(4, 1)}{C(9, 4)} = \frac{40}{126} = \frac{20}{63}$$

C - Among the fallen insects, there are exactly 2 ladybugs. This means we chose 2 ladybugs out of 4 and 2 beetles out of

5. The number of ways to do this is:

$$C(4, 2) \times C(5, 2) = \frac{4!}{2!2!} \times \frac{5!}{2!3!} = \frac{4 \times 3}{2 \times 1} \times \frac{5 \times 4}{2 \times 1} = 6 \times 10 = 60$$

The probability of event C is:

$$P(C) = \frac{C(4, 2) \times C(5, 2)}{C(9, 4)} = \frac{60}{126} = \frac{10}{21}$$

3. Final Answer

A - The probability that only Colorado beetles fell out is $\frac{5}{126}$.

B - The probability that 3 Colorado beetles and 1 ladybug fell out is $\frac{20}{63}$.

C - The probability that exactly 2 ladybugs fell out is $\frac{10}{21}$.