Axioms of Probability in Statistics

Question

I am studying Statistical Inference where suppose that a pair of balanced dice is tossed. Let $E_x$ be the event that the sum of the two numbers obtained is $x$, $x=2,3,\dotso ,12$. How do I find $P(E_x)$, $x = 2, 3, \dotso, 12$. Would say $P(2)=1/36$, $P(3)=2/36$, $P(4)=3/36$,..., $P(12)=1/36$? I need more explanation on this. What if we let $A$ be the event that $x$ is divisible by $4$, B be the event that $x>9$, $C$ be the event that $x$ is not a prime number. How do I find $P(A)$, $P(B)$, $P(C)$, $P(A\cap\ B)$, $P(A\cap\ C)$, $P(B\cap\ C)$, $P(A\cap\ B\cap\ C)$, $P(A\cup\ B \cup\ C)$, $P(A\cup\ B |C)$ and $P(A |B\cup\ C^c)$ ?

Answer 1

You can find $P(E_x)$ for $x=2,\dotso, 12$ by noting all the possible tosses which lead to the sum $x$.

For $x=2$, there is only the toss $(1,1)$, which gets you the sum $2$. For $x=3$, there is $(1,2)$ and $(2,1)$. And so on. Note, that there is some kind of symmetry in this. Of course $x=7$ has the highest probability, where

$(1,6), (2,5), (3, 4), (4,3), (5,2)$ and $(6,1)$ gets you the sum $7$.

Also note, that the probability for a specific toss (with two dice) is $\frac1{36}$.

So we have a Laplace experiment and the probability is calculated by:

$P(E_x)=\frac{\text{Number of tosses that lead to the sum x}}{\text{Number of all tosses}}=\frac{\text{Number of tosses that lead to the sum x}}{36}$

The other tasks can be tackled the same way. Just give the sets which are asked for.

For example: $A$ is the set of all $x$ (sum of the two dice) is divisible by $4$.

$x\in\{2,3,4,5,6,7,8,9,11,12\}$ so there are only three $x$ that are divisable by $4$.

Now you have to add the probabilities.