When two balanced dice are rolled, 36 equally likely outcomes are possible as shown below.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Let X denote the absolute value of the difference of the two numbers. Find the probability distribution of X. Give the probabilities as decimals rounded to three decimal places.
For X=0, the two dice must roll equal values, such as (1,1),(2,2), ....
Count the number of such cases, divide it by 36 (number of possible outcomes) and assign the quotient to X(0).
Repeat the count for X(1) (i.e. difference =1, such as (3,2)...), X(2)....until X(5).
The sum of X(0)...X(5) should add up to 1.
Make a frequency distribution for the absolute differences of the values. The absolute difference probabilities will be the number of tosses with each difference divided by 36. For example, no difference ([1,1], [2,2,] etc.) would be 6/36 = 1/6 (convert to decimal). I5I (absolute difference) would be 2/36.
I'll let you do the rest.
The probabilities should sum as 1 (within rounding error) as a check.
I hope this helps.
456589
To find the probability distribution of X, we need to determine the probability of each possible value of X occurring.
X can take values from 0 to 5, as the difference between two numbers on dice can range from 0 to 5.
Let's calculate the probability for each possible value of X:
For X = 0:
The only way to get X = 0 is when both dice show the same number. There are 6 such outcomes: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6). Therefore, the probability of X = 0 is 6/36 = 1/6 ≈ 0.167.
For X = 1:
To get X = 1, one die must show 1 more than the other die. There are 10 possible outcomes for this: (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (1, 5), (5, 1), (1, 6), and (6, 1). Therefore, the probability of X = 1 is 10/36 ≈ 0.278.
For X = 2:
To get X = 2, one die must show 2 more than the other die. There are 8 possible outcomes for this: (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), and (6, 4). Therefore, the probability of X = 2 is 8/36 ≈ 0.222.
For X = 3:
To get X = 3, one die must show 3 more than the other die. There are 6 possible outcomes for this: (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), and (6, 3). Therefore, the probability of X = 3 is 6/36 ≈ 0.167.
For X = 4:
To get X = 4, one die must show 4 more than the other die. There are 4 possible outcomes for this: (1, 5), (5, 1), (2, 6), and (6, 2). Therefore, the probability of X = 4 is 4/36 ≈ 0.111.
For X = 5:
To get X = 5, one die must show 5 more than the other die. There are 2 possible outcomes for this: (1, 6) and (6, 1). Therefore, the probability of X = 5 is 2/36 ≈ 0.056.
So, the probability distribution of X is:
X = 0: 1/6 ≈ 0.167
X = 1: 10/36 ≈ 0.278
X = 2: 8/36 ≈ 0.222
X = 3: 6/36 ≈ 0.167
X = 4: 4/36 ≈ 0.111
X = 5: 2/36 ≈ 0.056