# 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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