We have a jar containing 5 Colorado beetles and 4 ladybugs. Four insects fall out of the jar. We need to find the probabilities of the following events: A - only Colorado beetles fell out. B - 3 Colorado beetles and 1 ladybug fell out. C - exactly 2 ladybugs fell out.

1. Problem Description

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

A - only Colorado beetles fell out.

B - 3 Colorado beetles and 1 ladybug fell out.

C - exactly 2 ladybugs fell out.

2. Solution Steps

Total number of insects in the jar = 5 (beetles) + 4 (ladybugs) = 9

Number of insects that fell out = 4

The total number of ways to choose 4 insects from 9 is given by the combination formula:

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

Total number of ways to choose 4 insects from 9:

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

A - only Colorado beetles fell out:

We need to choose 4 beetles from the 5 beetles.

Number of ways to choose 4 beetles from 5:

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

Probability of event A:

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

B - 3 Colorado beetles and 1 ladybug fell out:

We need to choose 3 beetles from 5 and 1 ladybug from

4. Number of ways to choose 3 beetles from 5:

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

Number of ways to choose 1 ladybug from 4:

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

Number of ways to choose 3 beetles and 1 ladybug:

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

Probability of event B:

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

C - exactly 2 ladybugs fell out:

If 2 ladybugs fell out, then 2 beetles must have fallen out.

We need to choose 2 ladybugs from 4 and 2 beetles from

5. Number of ways to choose 2 ladybugs from 4:

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

Number of ways to choose 2 beetles from 5:

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

Number of ways to choose 2 ladybugs and 2 beetles:

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

Probability of event C:

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

3. Final Answer

A - $P(A) = \frac{5}{126}$

B - $P(B) = \frac{20}{63}$

C - $P(C) = \frac{10}{21}$