Question 1191527: MAT 145: Topics In Contemporary Math
Probability
3) Suppose we roll a pair of dice and add the numbers that land upmost. Fill in the table to
find the sample space.
1 2 3 4 5 6
1
2
3
4
5
6
Now find each of the following probabilities.
P(6) P(9) P(at least 9) P(doubles) P(at most 5)
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's say we had a red cube and a blue cube, each numbered 1 through 6.
This is what the list of all possible dice sums would be
| + | 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 |
For example, in the top left corner we have 1+1 = 2
Then just to the right of that we have 1+2 = 3
Followed by 1+3 = 4
and so on.
Now count how many times a black "6" shows up. It shows up exactly 5 times
Here are the five ways to sum to 6
1+5 = 6
2+4 = 6
3+3 = 6
4+2 = 6
5+1 = 6
Therefore, P(6) = 5/36 since there are 6*6 = 36 outcomes.
P(9) = 4/36 = 1/9 for similar reasoning
This time we have four ways to sum to 9 out of 36 outcomes total.
Those four sums are:
3+6 = 9
4+5 = 9
5+4 = 9
6+3 = 9
-------------------------
P(at least 9) = P(9 or more)
P(at least 9) = P(9) + P(10) + P(11) + P(12)
P(at least 9) = 4/36 + 3/36 + 2/36 + 1/36
P(at least 9) = (4+3+2+1)/36
P(at least 9) = 10/36
P(at least 9) = 5/18
Doubles is when you roll two of the same value. Eg: a pair of 3's
There are 6 such possible outcomes (one per side of the cube)
P(doubles) = 6/36 = 1/6
P(at most 5) = P(5 or less)
P(at most 5) = P(5) + P(4) + P(3) + P(2)
P(at most 5) = 4/36 + 3/36 + 2/36 + 1/36
P(at most 5) = (4+3+2+1)/36
P(at most 5) = 10/36
P(at most 5) = 5/18
This is nearly the same steps and exactly the same answer as P(at least 9) because of symmetry.
--------------------------
Answers:
P(6) = 5/36
P(9) = 1/9
P(at least 9) = 5/18
P(doubles) = 1/6
P(at most 5) = 5/18
|
|
|
