A box contains five numbered balls (1,2,2,3 and 4). We will randomly select two balls from the box (without replacement).

(a) The outcome of interest is the number on each of the two balls we select. List the complete sample space of outcomes.

(b) What is the probability of selecting two even-numbered balls?

(c) What is the probability that the number on the second selected ball is greater than the number on the first selected ball?

(d) Let X be the absolute difference of the two numbers on the selected balls |X1 - X2|, where X1 is the number on the first selected ball and X2 is the second selected ball. Find the probability distribution of X.

(e) What is the probability that X = 2 if the number 4 is not selected?

(f) Find the expected value of X.

(g) Find the variance of X.

a) SS= {12,13,14,21,22,23,24,34,43}

b) Selecting two even numbers
SS= {22,24,42}
P(getting 22) = (2/5)(1/4)= 0.1
P(getting 24) = (2/5)(1/4)= 0.1
P(getting 42) = (1/5)(2/4)= 0.1
P(getting 2 even number balls) = P(22)+P(24)+P(42)= 0.1+0.1+0.1= 0.3

c) The number on the second selected ball is greater than that of the first selected ball
SS={12,13,14,23,24,34}
P(getting 12)= (1/5)(2/4)= 0.1
P(getting 13)= (1/5)(1/4)= 0.05
P(getting 14)= (1/5)(1/4)= 0.05
P(getting 23)= (2/5)(1/4)= 0.1
P(getting 24)= (2/5)(1/4)= 0.1
P(getting 34)= (1/5)(1/4)= 0.05
P(number on the 2nd selected is greater than 1st selected ball)
= P(12)+P(13)+P(14)+P(23)+P(24)+P(34) 0.1+0.05+0.05+0.1+0.1+0.05=0.9
this is what I have so far... I'm having issues with question d, my probabilities aren't adding up to 1

I am not certain what your question is here. Do you want your work checked?

15

(a) To list the complete sample space of outcomes, we need to consider all possible combinations of two balls selected from the box. Without replacement means that once a ball is selected, it cannot be selected again.

The possible outcomes are as follows:
1. (1, 2)
2. (1, 3)
3. (1, 4)
4. (2, 1)
5. (2, 3)
6. (2, 4)
7. (3, 1)
8. (3, 2)
9. (3, 4)
10. (4, 1)
11. (4, 2)
12. (4, 3)

(b) To find the probability of selecting two even-numbered balls, we need to find the number of favorable outcomes (2, 2, 4) and divide it by the total number of possible outcomes.

Favorable outcomes: 3 (2, 2, 4)
Total number of possible outcomes: 12

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = 3 / 12
Probability = 1 / 4 = 0.25

Therefore, the probability of selecting two even-numbered balls is 0.25 or 25%.

(c) To find the probability that the number on the second selected ball is greater than the number on the first selected ball, we need to count the number of favorable outcomes and divide it by the total number of possible outcomes.

Favorable outcomes: (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) (6 outcomes)
Total number of possible outcomes: 12

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = 6 / 12
Probability = 1 / 2 = 0.5

Therefore, the probability that the number on the second selected ball is greater than the number on the first selected ball is 0.5 or 50%.

(d) To find the probability distribution of X, we need to calculate the absolute difference for each pair of numbers and count the number of occurrences.

X = |X1 - X2|

Possible values of X and their probabilities:
X = 0: (2, 2) with probability 1/12 (1 occurrence)
X = 1: (2, 4) and (4, 2) with probability 2/12 (2 occurrences)
X = 2: (1, 3), (1, 4), (3, 1), (3, 2), (4, 1), (4, 3) with probability 6/12 (6 occurrences)

Therefore, the probability distribution of X is:
X = 0 with probability 1/12
X = 1 with probability 2/12
X = 2 with probability 6/12

(e) To find the probability that X = 2 if the number 4 is not selected, we need to consider the outcomes from the sample space where the number 4 is not selected.

Outcomes without the number 4: (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)

Number of favorable outcomes: 0 (none of the outcomes have X = 2)
Total number of possible outcomes: 6

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = 0 / 6
Probability = 0

Therefore, the probability that X = 2 if the number 4 is not selected is 0.

(f) To find the expected value of X, we need to multiply each possible value of X by its corresponding probability and sum them up.

Expected value of X = (0 * 1/12) + (1 * 2/12) + (2 * 6/12)
Expected value of X = 0 + 1/6 + 1
Expected value of X = 7/6

Therefore, the expected value of X is 7/6 or approximately 1.167.

(g) To find the variance of X, we need to calculate the squared difference between each possible value of X and the expected value of X, multiply it by its probability, and sum them up.

Variance of X = [(0 - 7/6)^2 * 1/12] + [(1 - 7/6)^2 * 2/12] + [(2 - 7/6)^2 * 6/12]

Variance of X = [(0 - 7/6)^2] * 1/12 + [(1 - 7/6)^2] * 2/12 + [(2 - 7/6)^2] * 6/12

Variance of X = (49/36) * 1/12 + (1/36) * 2/12 + (1/36) * 6/12

Variance of X = 49/432 + 2/432 + 6/432

Variance of X = 57/432

Therefore, the variance of X is 57/432 or approximately 0.132.