Let X be a random variable that records the outcome of a dice toss. The probability distribution for X is as follows:
Outcome: 1, 2, 3, 4, 5, 6
Probability: 1/6, 1/6, 1/6, 1/6, 1/6, 1/6
Suppose now that we toss two dice. X1 records the outcome of the first toss, and X2 records the outcome of the second toss. We are interested in the sum of the two dice tosses,
Y= X1+ X2.
What is the mean value of Y, μY?
To find the mean value of Y, μY, we need to sum up the values of Y multiplied by their respective probabilities.
The possible outcomes for Y are:
2: (1, 1)
3: (1, 2), (2, 1)
4: (1, 3), (2, 2), (3, 1)
5: (1, 4), (2, 3), (3, 2), (4, 1)
6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
9: (3, 6), (4, 5), (5, 4), (6, 3)
10: (4, 6), (5, 5), (6, 4)
11: (5, 6), (6, 5)
12: (6, 6)
We can calculate the value of Y for each outcome and also calculate the probability for each outcome by multiplying the probabilities of each individual dice toss. Then, we multiply Y by its probability and sum up these products to find μY.
μY = (2 * P(2)) + (3 * P(3)) + (4 * P(4)) + (5 * P(5)) + (6 * P(6)) + (7 * P(7)) + (8 * P(8)) + (9 * P(9)) + (10 * P(10)) + (11 * P(11)) + (12 * P(12))
Using the given probability distribution for X, we can calculate μY as follows:
μY = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + (5 * 4/36) + (6 * 5/36) + (7 * 6/36) + (8 * 5/36) + (9 * 4/36) + (10 * 3/36) + (11 * 2/36) + (12 * 1/36)
Simplifying this equation:
μY = 2/36 + 6/36 + 12/36 + 20/36 + 30/36 + 42/36 + 40/36 + 36/36 + 30/36 + 22/36 + 12/36
μY = 244/36
μY ≈ 6.7778
Therefore, the mean value of Y, μY, is approximately 6.7778.
To find the mean value (μY) of Y, we need to calculate the expected value or average.
In this case, we have two dice tosses, and Y represents the sum of the outcomes of those tosses (Y = X1 + X2). To find the mean value of Y, we need to calculate the expected value of Y.
To calculate the expected value, we multiply each possible value of Y by its corresponding probability and sum up the results.
Let's calculate it step by step:
1. Determine all possible outcomes for Y:
- Y can be any number from 2 to 12 (since the minimum sum of two dice is 1+1 = 2, and the maximum is 6+6 = 12).
2. Calculate the probability of each possible outcome:
- To find the probability of each sum, we need to consider all the possible combinations of the outcomes of X1 and X2 that yield that sum. For example, to achieve a sum of 2, we can have (1, 1) as the outcomes of X1 and X2.
The possible sums and their corresponding probabilities are as follows:
- Sum of 2: The only possible combination is (1, 1) with a probability of (1/6 * 1/6) = 1/36.
- Sum of 3: Possible combinations are (1, 2) and (2, 1), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 4: Possible combinations are (1, 3), (2, 2), and (3, 1), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 5: Possible combinations are (1, 4), (2, 3), (3, 2), and (4, 1), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 6: Possible combinations are (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 7: Possible combinations are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 8: Possible combinations are (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 9: Possible combinations are (3, 6), (4, 5), (5, 4), and (6, 3), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 10: Possible combinations are (4, 6), (5, 5), and (6, 4), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 11: Possible combinations are (5, 6) and (6, 5), each with a probability of (1/6 * 1/6) = 1/36.
- Sum of 12: The only possible combination is (6, 6) with a probability of (1/6 * 1/6) = 1/36.
3. Calculate the expected value (mean) of Y:
- Now, we multiply each possible sum by its corresponding probability and sum up the results:
(2 * (1/36)) + (3 * (1/36)) + (4 * (1/36)) + (5 * (1/36)) + (6 * (1/36)) + (7 * (1/36)) + (8 * (1/36)) + (9 * (1/36)) + (10 * (1/36)) + (11 * (1/36)) + (12 * (1/36))
Simplifying the expression, we get:
μY = (1/36) * (2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)
Calculating this, we find:
μY = (1/36) * 77
= 77/36
≈ 2.1389
Therefore, the mean value (μY) of Y, which represents the sum of two dice tosses with the given probability distribution, is approximately 2.1389.
To find the mean value of Y, we need to calculate the expected value of the sum of the two dice tosses.
The expected value of a random variable can be found by multiplying each outcome by its corresponding probability and summing them up.
In this case, we have the following possible outcomes for the sum of two dice tosses:
Outcome: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
To calculate the expected value, we multiply each outcome by its probability and sum them up:
E(Y) = 2 * P(Y=2) + 3 * P(Y=3) + 4 * P(Y=4) + 5 * P(Y=5) + 6 * P(Y=6) + 7 * P(Y=7) + 8 * P(Y=8) + 9 * P(Y=9) + 10 * P(Y=10) + 11 * P(Y=11) + 12 * P(Y=12)
Plugging in the corresponding probabilities from the given probability distribution, we get:
E(Y) = 2 * (1/36) + 3 * (2/36) + 4 * (3/36) + 5 * (4/36) + 6 * (5/36) + 7 * (6/36) + 8 * (5/36) + 9 * (4/36) + 10 * (3/36) + 11 * (2/36) + 12 * (1/36)
E(Y) = (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) / 36
E(Y) = 252 / 36
E(Y) = 7
Therefore, the mean value of Y, μY, is 7.