SOLUTION: q1)
A) random variable X is de fined as the larger of the scores obtained in two throws of a fair six-sided die.Find the distribution of the random variable X.
(b)A random varia
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Question 1197519: q1)
A) random variable X is de fined as the larger of the scores obtained in two throws of a fair six-sided die.Find the distribution of the random variable X.
(b)A random variable Y is de fined as the highest score obtained in k throws of a fair six-sided die.Determine the probability mass function of Y .
q2)
.A white die and a red die are thrown at the same time and the difference W−R is observed ,where R is the number on top of the red die and W is that on top of the white one .Find the probability mass function of this difference W−R.
Answer by math_tutor2020(3835) (Show Source): You can put this solution on YOUR website!
Question 1, Part (a)
This is one way to write out the table of possible outcomes.
| 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 2 | 3 | 4 | 5 | 6 |
| 3 | 3 | 3 | 3 | 4 | 5 | 6 |
| 4 | 4 | 4 | 4 | 4 | 5 | 6 |
| 5 | 5 | 5 | 5 | 5 | 5 | 6 |
| 6 | 6 | 6 | 6 | 6 | 6 | 6 |
Example: We roll a 6 on the blue die and a 1 on the red die. The result is 6 since it's the larger of the two outcomes (top right corner of the table).
The possible outcomes are: 1,2,3,4,5,6
Let X be those possible outcomes.
Here's the frequency chart
We'll divide each frequency over 36 since there are 6*6 = 36 outcomes.
Therefore, we have this probability mass function (PMF).
| X | P(X) |
| 1 | 1/36 |
| 2 | 3/36 |
| 3 | 5/36 |
| 4 | 7/36 |
| 5 | 9/36 |
| 6 | 11/36 |
I decided not to reduce the fractions so that each could keep the consistent denominator of 36.
If you want to reduce the fractions, then,
3/36 = 1/12
9/36 = 1/4
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Question 1, Part (b)
Right now I'm blanking on how to do this, so I'll come back to this later. Or I'll let another tutor step in. Sorry for the trouble.
The slight good news is that I managed to find the PMF tables for k = 3 through k = 5 using computer software. But I wasn't able to find a generalized case for any positive integer k value.
PMF for k = 3
| X | P(X) |
| 1 | 1/216 |
| 2 | 7/216 |
| 3 | 19/216 |
| 4 | 37/216 |
| 5 | 61/216 |
| 6 | 91/216 |
PMF for k = 4
| X | P(X) |
| 1 | 1/1296 |
| 2 | 15/1296 |
| 3 | 65/1296 |
| 4 | 175/1296 |
| 5 | 369/1296 |
| 6 | 671/1296 |
PMF for k = 5
| X | P(X) |
| 1 | 1/7776 |
| 2 | 31/7776 |
| 3 | 211/7776 |
| 4 | 781/7776 |
| 5 | 2101/7776 |
| 6 | 4651/7776 |
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Question 2
We'll use the template from the table in question 1, part (a).
Instead of white, I'll use blue.
We subtract the values in the format B - R
B = blue
R = red
| 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | -1 | 0 | 1 | 2 | 3 | 4 |
| 3 | -2 | -1 | 0 | 1 | 2 | 3 |
| 4 | -3 | -2 | -1 | 0 | 1 | 2 |
| 5 | -4 | -3 | -2 | -1 | 0 | 1 |
| 6 | -5 | -4 | -3 | -2 | -1 | 0 |
For example, if we roll a 6 on the blue die and a 1 on the red die, then B-R = 6-1 = 5 which is in the top right corner of the table.
The outcomes range from -5 to +5 inclusive of the endpoints.
Let X be the result of each difference
| X | Frequency |
| -5 | 1 |
| -4 | 2 |
| -3 | 3 |
| -2 | 4 |
| -1 | 5 |
| 0 | 6 |
| 1 | 5 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 1 |
Then we divide each frequency over 36 to form the PMF.
| X | P(X) |
| -5 | 1/36 |
| -4 | 2/36 |
| -3 | 3/36 |
| -2 | 4/36 |
| -1 | 5/36 |
| 0 | 6/36 |
| 1 | 5/36 |
| 2 | 4/36 |
| 3 | 3/36 |
| 4 | 2/36 |
| 5 | 1/36 |
Once again, I chose not to reduce the fractions to keep the same denominator (36).
If you want to reduce the fractions, then,
2/36 = 1/18
3/36 = 1/12
4/36 = 1/9
6/36 = 1/6
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