3. Probability of events

Probability of events

Event

An event is a subset of the [[Sample space|sample space]], or otherwise a set of possible outcomes.

$$P(A)=\sum _{\{\omega \in A\}}P(\omega )$$ Link to original

Probability of an event

The probability of an event $E$ is the sum of the probabilities of the outcomes in $E$.

title: Example 1
If two fair dice are rolled, what is the probability that the sum is $7$?
 
By product rule, there are $6 \cdot 6 = 36$ possible outcomes, each of them equally likely with a probability of $\frac{1}{36}$.
 
There are $6$ successful outcomes: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$.
 
Hence probability is: $6 \cdot \frac{1}{36} = \boxed{\frac{1}{6}}$

title: Example 2
Given that a dice is weighed in such a way that $3$ appaers twice as often as each of the other numbers but the others are equally likely. What is the probability we roll an odd number?
 
We need to find $p(E)$ of the event $E = \{ 1, 3, 5 \}$.
 
We know that:
- $\sum^n_1 p(n) = 1$
- $p(1) = p(2) = p(4) = p(5) = p(6)$
- $p(3) = 2p(1)$
 
Hence $p(1) = p(2) = p(4) = p(5) = p(6) = \frac{1}{7}$ and $p(3) = \frac{2}{7}$.
 
So, $p(E) = p(1) + p(3) + p(5) = \frac{4}{7}$.

title: Example 3
Take for example a lottery, where we can win if we correctly choose a set of $6$ numbers out of the first $59$ positive natural numbers.
 
The total number of ways to choose $6$ numbers out of $59$ is found using [[Order does not matter, repetition not allowed|order does not matter, repetition not allowed]] combinations:
 
$$
	\begin{pmatrix} 59 \\ 6 \end{pmatrix}
	= \frac{59 \cdot 58 \cdot 57 \cdot 56 \cdot 55 \cdot 54}
	{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}
	= 45, 057, 474
$$
 
Hence we find that:
 
$$
	p(W) = \frac{1}{45, 057, 474}
$$

Link to original