Problems from Laplace's rule

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 $$\mathrm{\Omega }=\{1-1,1-2,1-3,1-4,1-5,1-6,2-1,2-2,\dots \}$$

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 $6\cdot 6=36$ possible ways.

All cases are equiprobable, and therefore the probability of each elementary event is ${\displaystyle \frac{1}{36}}$, 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,

$P(A_1)={\displaystyle \frac{4}{36}}$, $P(A_2)={\displaystyle \frac{3}{36}}$, $P(A_3)={\displaystyle \frac{2}{36}}$, $P(A_4)={\displaystyle \frac{1}{36}}$.

Therefore $$P(A)=P(A_1\cup A_2\cup A_3\cup A_4)=$$ $$={\displaystyle \frac{4}{36}}+{\displaystyle \frac{3}{36}}+{\displaystyle \frac{2}{36}}+{\displaystyle \frac{1}{36}}={\displaystyle \frac{11}{36}}$$

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:

$B_1=\mathrm{\varnothing }$, 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, $$P(B)=P(B_1\cup B_2\cup B_3)=$$ $$=0+{\displaystyle \frac{1}{36}}+{\displaystyle \frac{2}{36}}={\displaystyle \frac{3}{36}}={\displaystyle \frac{1}{12}}.$$

As $B_1$ is an impossible event, it has no favorable outcome, and therefore, in accordance with Laplace's rule, its probability is $P(B_1)={\displaystyle \frac{0}{36}}=0$.

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 $$P(C)=P(A\cup B)=P(A)+P(B)={\displaystyle \frac{11}{36}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{14}{36}}={\displaystyle \frac{7}{18}}.$$

Solution:

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

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

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

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