Question 1193853: 2 ordinary dice are thrown find the probability that
1. The sum of the two dice is 3
2. The sum of the two dice exceeds 9
3. The two dice show the same number
4. The number on the two dice is differ by more than 2
5. The product of the two dice is even
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
You posted a lot of questions here. Please post one question at a time.
I'll do the first three questions to get you started.
Problem 1
Here's a sum of dice chart
+ | 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 |
Imagine we had a red die and blue die. The table above shows all the possible sums. For instance, 2+3 = 5 in row 2, column 3. The rest of the table is filled out the same way.
As a quick way to fill out the table without doing much addition, notice how we have 2,3,4,5,6,7 in the first row, then 3,4,5,6,7,8 in the second row and so on. The numbers increment by 1, and we have this shift going on.
Once the table is set up, it should be fairly easy to answer each question.
Based on the table, we have these possible ways to get a sum of 3
1+2 = 3
2+1 = 3
This is out of 6*6 = 36 ways to roll two six-sided dice.
Therefore, the probability of getting a sum of 3 is 2/36 = 1/18
Answer: 1/18
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Problem 2
Refer to the dice chart in problem 1.
We have these 6 ways to get a sum larger than 9
4+6 = 10
5+5 = 10
6+4 = 10
5+6 = 11
6+5 = 11
6+6 = 12
Therefore, the probability is 6/36 = 1/6
Answer: 1/6
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Problem 3
You can refer to the chart above if you want, or you can list out the cases like so
1+1 = 2
2+2 = 4
3+3 = 6
4+4 = 8
5+5 = 10
6+6 = 12
All of which are along the main diagonal that points to the northwest.
Put another way, we have 6 cases here because each of the six faces has a corresponding pair.
There are 6 ways to get the dice to have their numbers match up out of 36 ways to roll the dice.
6/36 = 1/6
Answer: 1/6
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