# Problems from Laplace's rule

We throw two six-sided dice. Adding the two results, we want to know the probability of:

a) $$A =$$ "get a total higher than $$8$$"

b) $$B =$$ "a total less than or equal to $$3$$"

c) $$C =$$ "a total higher than $$8$$, or less than or equal to $$3$$"

See development and solution

### Development:

First, we have to describe our sample space. In each dice there may came out a number between $$1$$ and $$6$$, so that all possible outcomes are $$\Omega=\{1-1,1-2,1-3,1-4,1-5,1-6,2-1,2-2,\ldots\}$$$Note that $$36$$ items would be: we can see it thinking that in the first dice it can come out a number between $$1$$ and $$6$$, and in the second one the same, so in total there are $$6\cdot6=36$$ possible ways. All cases are equiprobable, and therefore the probability of each elementary event is $$\dfrac{1}{36}$$, in virtue of Laplace's rule. a) We want to see which ones are the results favorable to $$A =$$ "get more than $$8$$." The results that will meet are all those that turn out to be $$9, 10, 11,$$ and $$12$$. There cannot be results that are higher since the maximum when throwing both dices, if the both of them turn to be $$6$$, is $$6+6=12$$. So, we are saying that, $$A=A_1 \cup A_2 \cup A_3 \cup A_4$$, and we have to think about what events accomplish the following: $$A_1=$$"total addition $$=9$$"$$=\{3-6,4-5,5-4,6-3\}$$, that is, there are four favorable cases. $$A_2=$$"total addition $$=10$$"$$=\{4-6,5-5,6-4\}$$, three favorable cases. $$A_3=$$"total addition $$=11$$"$$=\{5-6,6-5\}$$, two favorable cases. $$A_4=$$"total addition $$=12$$"$$=\{6-6\}$$, one favorable case. As all the events are equiprobable, we can apply Laplace's rule, $$P(A_1)=\dfrac{4}{36}$$, $$P(A_2)=\dfrac{3}{36}$$, $$P(A_3)=\dfrac{2}{36}$$, $$P(A_4)=\dfrac{1}{36}$$. Therefore $$P(A)=P(A_1 \cup A_2 \cup A_3 \cup A_4)=$$$ $$=\dfrac{4}{36}+\dfrac{3}{36}+\dfrac{2}{36}+\dfrac{1}{36}=\dfrac{11}{36}$$$b) To calculate $$P(B)$$, we do as we have done before. Let $$B_1=$$"add up $$1$$", $$B_2=$$"add up $$2$$", $$B_3=$$"add up $$3$$". Then the possible cases that satisfy each of these events are: $$B_1=\emptyset$$, because it is an impossible event. The smallest result that we could obtain would be $$2$$, that is, one on both dices, $$1+1=2$$. $$B_2=\{1-1\}$$ $$B_3=\{1-2,2-1\}$$ So, $$P(B)=P(B_1 \cup B_2 \cup B_3)=$$$ $$=0+\dfrac{1}{36}+\dfrac{2}{36}=\dfrac{3}{36}=\dfrac{1}{12}.$$$As $$B_1$$ is an impossible event, it has no favorable outcome, and therefore, in accordance with Laplace's rule, its probability is $$P(B_1)=\dfrac{0}{36}=0$$. c) In this case, we can see all the events that satisfy $$C$$, but it's a long procedure. We can solve it easily if we note that $$C=A \cup B$$. We can calculate $$P(C)=P(A \cup B)=P(A)+P(B)=\dfrac{11}{36}+\dfrac{1}{12}=\dfrac{14}{36}=\dfrac{7}{18}.$$$

### Solution:

a) $$P(A)=\dfrac{11}{36}$$

b) $$P(B)=\dfrac{1}{12}$$

c) $$P(C)=\dfrac{7}{18}$$

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