A box contains four balls labeled 1,2,3,4 respectively. Two balls will be sequentially selected from the box at random without replacement. Let Y1 be the maximum of two outcomes and Y2 be the minimum of two outcomes observed. Find the joint probability function of Y1 and Y2.

To find the joint probability function of Y1 and Y2, we need to consider all possible outcomes and calculate the probability for each combination.

Let's analyze the possible values for Y1 and Y2:

Y1 represents the maximum value of the two outcomes. The maximum value can only be one of the four numbers: 1, 2, 3, or 4.
Y2 represents the minimum value of the two outcomes. The minimum value can also only be one of the four numbers: 1, 2, 3, or 4.

To find the joint probability function, we need to determine the probability of each combination of Y1 and Y2. There are four possibilities for Y1 and Y2 in this case:

1. If Y1 and Y2 are both 1:
The probability of this event is the probability of selecting the ball labeled 1 twice in a row, without replacement. The probability of selecting ball 1 on the first draw is 1/4. After removing the selected ball, the probability of selecting it again is 0 since it is no longer in the box. Therefore, the joint probability for Y1=1 and Y2=1 is (1/4)*(0) = 0.

2. If Y1=1 and Y2=2:
The probability of this event is the probability of selecting the ball labeled 1 first and the ball labeled 2 second, without replacement. The probability of selecting ball 1 on the first draw is 1/4. After removing the selected ball, the probability of selecting ball 2 is 1/3. Therefore, the joint probability for Y1=1 and Y2=2 is (1/4)*(1/3) = 1/12.

3. If Y1=2 and Y2=2:
The probability of this event is the probability of selecting the ball labeled 2 twice in a row, without replacement. The probability of selecting ball 2 on the first draw is 1/4. After removing the selected ball, the probability of selecting it again is 0 since it is no longer in the box. Therefore, the joint probability for Y1=2 and Y2=2 is (1/4)*(0) = 0.

4. If Y1=3 and Y2=4:
The probability of this event is the probability of selecting the ball labeled 4 first and the ball labeled 3 second, without replacement. The probability of selecting ball 4 on the first draw is 1/4. After removing the selected ball, the probability of selecting ball 3 is 1/3. Therefore, the joint probability for Y1=3 and Y2=4 is (1/4)*(1/3) = 1/12.

Since the joint probability function represents the probability of each combination of Y1 and Y2, we can summarize the results in a table:

Y1 | Y2 | Probability
-------------------------
1 | 1 | 0
1 | 2 | 1/12
2 | 2 | 0
3 | 4 | 1/12

Therefore, the joint probability function of Y1 and Y2 is given by:

P(Y1=1, Y2=1) = 0
P(Y1=1, Y2=2) = 1/12
P(Y1=2, Y2=2) = 0
P(Y1=3, Y2=4) = 1/12