Hi Chiluya,
I problems concerning rolling two dice I find it useful to think about the 6 by 6 array of possible outcomes shown below. I am going to think of the green die as the first one tossed and the red die as the second. Each of the 36 cells in the array represents an outcome of the pair of dice. For example the cell containing $1, 5$ represents the outcome of a $1$ on the first die and a $5$ on the second die. Each of the 36 possible outcomes is equally likely on a toss of the dice. Fill in the rest of the cells.
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1 |
2 |
3 |
4 |
5 |
6 |
1 |
1,1 |
1,2 |
1,3 |
1,4 |
1,5 |
1,6 |
2 |
2,1 |
2,2 |
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3 |
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4 |
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5 |
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6 |
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I want to show you two methods to approach this problem.
Method 1
You have tossed the dice and you know the sum of the two dice was 8. Thus you know you are in one of 5 cells in the table above. Can you see which 5? Each of these 5 cells are equally likely. How many of them have the first die as least as large as 4? What is the probability that the first die is at least as large as 4 if you know that the sum is 8?
Method 2
This method requires that you know that the conditional probability of an event $A$ given$B$ is
\[Pr[A|B] = \frac{Pr(A \mbox{ and } B)}{Pr(B)}.\]
In your case $A$ is the event that the result on the first die is at least as large as 4 and $B$ is the event that the sum of the two dice is 8. Look at the table above that you completed. How many cells have the sum of the two dice as 8? What is the probability that when you roll the two dice the sum of two dice is 8? Again from the table how many of the cells have the first die as large as 4 AND the sum of the two dice is 8? What is $Pr[A \mbox{ and } B]?$ What is $Pr[A|B]?$
Penny
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