There are 5 Colorado beetles and 4 ladybugs in a jar. Four insects fall out of the jar. We need to find the probabilities of the following events: A - only Colorado beetles fall out; B - 3 Colorado beetles and 1 ladybug fall out; C - exactly 2 ladybugs are among the fallen insects.

1. Problem Description

There are 5 Colorado beetles and 4 ladybugs in a jar. Four insects fall out of the jar. We need to find the probabilities of the following events:

A - only Colorado beetles fall out;

B - 3 Colorado beetles and 1 ladybug fall out;

C - exactly 2 ladybugs are among the fallen insects.

2. Solution Steps

First, we need to find the total number of ways to choose 4 insects from the 9 insects in the jar. This is given by the combination formula:

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

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 fall out.

We need to choose 4 Colorado beetles out of

5. $C(5, 4) = \frac{5!}{4!1!} = 5$

The probability of this event is:

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

B - 3 Colorado beetles and 1 ladybug fall out.

We need to choose 3 Colorado beetles out of 5 and 1 ladybug out of

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

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

The number of ways to choose 3 Colorado beetles and 1 ladybug is:

$C(5, 3) \times C(4, 1) = 10 \times 4 = 40$

The probability of this event is:

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

C - Exactly 2 ladybugs are among the fallen insects.

This means we need to choose 2 ladybugs out of 4 and 2 Colorado beetles out of

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

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

The number of ways to choose 2 ladybugs and 2 Colorado beetles is:

$C(4, 2) \times C(5, 2) = 6 \times 10 = 60$

The probability of this event 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 fall out is $\frac{5}{126}$.

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

C - The probability that exactly 2 ladybugs are among the fallen insects is $\frac{10}{21}$.