We have a box with four numbered clips: 1, 2, 3, and 4. Two clips are drawn one at a time without replacement. We need to find the probability of the following events: A. One clip has an odd number, and the other clip has an even number. B. The numbers on both clips are multiples of 2. C. The sum of the numbers on both clips is a multiple of 3.

1. Problem Description

We have a box with four numbered clips: 1, 2, 3, and

4. Two clips are drawn one at a time without replacement. We need to find the probability of the following events:

A. One clip has an odd number, and the other clip has an even number.

B. The numbers on both clips are multiples of

2. C. The sum of the numbers on both clips is a multiple of

3.

2. Solution Steps

First, we need to find the total number of possible outcomes when drawing two clips without replacement. This can be calculated as permutations. The total number of permutations is $4 \times 3 = 12$.

A. One clip has an odd number, and the other has an even number.

Possible outcomes: (1,2), (1,4), (2,1), (2,3), (3,2), (3,4), (4,1), (4,3).

There are 8 such outcomes.

The probability of A is $\frac{8}{12} = \frac{2}{3}$.

B. The numbers on both clips are multiples of

2. The multiples of 2 are 2 and

4. Possible outcomes: (2,4), (4,2).

There are 2 such outcomes.

The probability of B is $\frac{2}{12} = \frac{1}{6}$.

C. The sum of the numbers on both clips is a multiple of

3. Possible outcomes:

1+2 = 3

1+5 (impossible)

2+1 = 3

2+4 = 6

3+3 (impossible)

3+6 (impossible)

4+2 = 6

4+5 (impossible)

Possible pairs: (1,2), (2,1), (2,4), (4,2).

3+6 (impossible)

4+5(impossible)

(1,2): sum is 3

(2,1): sum is 3

(2,4): sum is 6

(4,2): sum is 6

So the pairs are (1,2), (2,1), (2,4), (4,2).

There are 4 such outcomes.

The probability of C is $\frac{4}{12} = \frac{1}{3}$.

3. Final Answer

A: The probability that one clip has an odd number and the other clip has an even number is $\frac{2}{3}$.

B: The probability that the numbers on both clips are multiples of 2 is $\frac{1}{6}$.

C: The probability that the sum of the numbers on both clips is a multiple of 3 is $\frac{1}{3}$.