Question 1197173: When two balanced dice are rolled, 36 equally likely outcomes are possible. Let X denote the sum of the dice. a) What are the possible values of the random variable X ? b) Use random-variable notation to represent the event that the sum of the dice is 7. c) Find P(X = 7). d) Find the probability distribution of X . Leave your probabilities in fraction form. e) Construct a probability histogram for X.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
X = sum of the two dice
Assuming we have two standard six-sided dice, each labeled 1 through 6, then the possible values of X are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
due to this sum of dice chart here
| + | 1 | 2 | 3 | 4 | 5 | 6 | | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 | | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Each value in black is a sum of a red and blue value.
Eg: 1 + 6 = 7 in the top right corner.
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Part (b)
Here are all the possible ways to get a sum of 7
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
I'm not exactly sure what your teacher is referring to when mentioning "Use random-variable notation".
It's possible that your teacher wants you to write P(X = 7) for the answer here.
I recommend getting a second opinion from your teacher, another classmate, or another tutor.
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Part (c)
In the previous part, there were 6 ways to get a sum of 7
This is out of 36 total outcomes
6/36 = 1/6
P(X = 7) = 1/6
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Part (d)
We'll follow the same idea as part (c), but do so for X = 2 all the way up to X = 12
P(X = 2) = 1/36 since there is one way to get a sum of 2 out of 36 total outcomes
P(X = 3) = 2/36 since there are two ways to get a sum of 3 out of 36 total outcomes
P(X = 4) = 3/36 since there are three ways to get a sum of 4 out of 36 total outcomes
And so on until reaching P(X = 7) = 6/36
So far the numerators have increased: 1,2,3,4,5,6
Then we have a countdown of numerators: 5,4,3,2,1 when going through P(X = 8) to P(X = 12)
Here's what the probability distribution looks like
| X | P(X) | | 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 |
I'll leave the fractions un-reduced so that each has a denominator of 36. This way things remain consistent and you can compare each P(X) value.
If your teacher requires you to reduce each fraction, then be sure to do so.
Here is each reduction
2/36 = 1/18
3/36 = 1/12
4/36 = 1/9
6/36 = 1/6
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Part (e)
Refer to the table in part (d)
Each P(X) value tells us how tall the bar will be for any particular X value.
Example: The first bar is 1/36 units tall since P(X) = 1/36 when X = 2.
This is what the histogram would look like
This distribution is symmetric since the left half is a mirror copy of the right half.
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