First let us understand what an 'event' means in terms of math or statistics. If we conduct an experiment, then there are many possible outcomes of this experiment. A set of all the possible outcomes of a probability experiment is called its sample space. An event is a subset of this sample space. It includes the outcomes that we are interested in or the outcomes with some specific characteristics. For example, if the experiment involved is about throwing a six sided fair die. Then there are $6$ possible outcomes. But if we are interested in finding the probability of a prime number showing up, then that becomes our event. It would be denoted as follows:

$A$ = $\{x|x$ is a prime number, $1 \leq x \leq 6 \}$

Usually we use capital alphabets to denote events.

The likelihood of occurrence of a specific event is called its probability. If we consider an event $E$, then the number of outcomes that favour event $E$ divided by the total number of outcomes in the sample space $(S)$ would be the probability of the event $E$. From the definition it is obvious that the number of elements in set $E$ can be either equal to or less than the number of elements in the sample space $S$. Thus the probability of an event has to be less than or equal to $1$. It can never be greater than $1$. Also, the number of elements in set $E$ cannot be smaller than zero. A set cannot have a negative cardinal number. At most it can be a null set with no elements. Thus the probability of an event has to be greater than or equal to $0$. Putting both these conditions together we can say that:

$0 \leq P \leq 1$

Where, $P$ is the probability of an event.

Consider that we have conducted an experiment. The total number of possible outcomes of this experiment is $n$. This $n$ is the cardinal number of the sample space $S$. Now out of these n possible outcomes, there are $r$ ways in which a particular event $E$ can occur. Then the probability of this event $E$ is denoted by $P(E)$ and it can be calculated using the formula:

$P(E)$ = $\frac{(\text{Number of favourable outcomes})}{(\text{Total number of possible outcomes})}$

$P(E)$ = $\frac{r}{n}$

This is the probability of occurrence of the event $E$. Now what if we want to find the probability of the event $E$ not occurring? Since we already established earlier that the total maximum probability is always $1$, the probability of non-occurrence of an event would be:

$P(E')$ = $1 - P(E)$

$P(E')$ = $1 - \frac{r}{n}$

The event $E'$ is called the complementary event of the event $E$. In other words we can state that:

$P(E) + P(E')$ = $1$

Consider a probability experiment that has n possible outcomes. If each of these outcomes are equally likely, then the probability of each of the outcomes would be:

$P$ = $\frac{1}{n}$

Now if we have an event $E$ that encompasses $r$ out of these n equally likely outcomes. Then the probability of the event $E$ would be:

$P(E)$ = $r$ $(\frac{1}{n})$ = $\frac{r}{n}$

If event $E$ is such that it includes all the members of the entire sample space $S$ then its probability would be:

$P(E)$ = $n$ $(\frac{1}{n})$ = $1$

Hence the total probability of $n$ equally likely events would be $1$.

To find the probability of a single event, we need to first find out the number of ways in which that event can occur. Then we divide that number by the total number of ways in which the experiment can occur. That would give us the required probability. We shall see examples of this in a while.

Conditional probability comes to play when we are talking of two events $A$ and $B$ that are part of the same sample space $S$. If the occurrence of one event depends on whether the other event has occurred or not, then the probability thus determined is called conditional probability. Thus the conditional probability of event $B$ given that event $A$ has already occurred is denoted by:

$P(B|A)$

The formula for calculating the conditional probability of a single event would be:

$P(B│A)$ = $\frac{P(A\ and\ B)}{P(A)}$

This formula can be used when both the events $A$ and $B$ are not independent of each other. Another way of stating this formula could be:

$P(A$ and $B)$ = $P(A) \times P(B|A)$

If we have three events, $A, B$ and $C$ then the probability of occurrence of all these three events together would be:

$P(A$ and $B$ and $C)$ = $P(A) \times P(B│A) \times P(C|A$ and $B)$

Find the probability of drawing a Queen from a standard deck of $52$ cards, if a card is drawn at random.

There are $52$ cards in the standard deck. So the total number of possible outcomes in the sample space is $52$.

The outcomes that we would be interested in are the ones that have Queen. There are $4$ Queens in a standard deck. Thus the number of favourable outcomes is $4$.

Therefore the probability can be written as:

$P(E)$ = $\frac{(\text{number of favourable outcomes})}{(\text{total number of possible outcomes})}$

$P(E)$ = $\frac{4}{52}$ = $\frac{1}{13}$ $\leftarrow\ Answer!$

A fair six sided die is rolled. Find the probability of a prime number showing up.

Since the die has $6$ sides, therefore there are $6$ possible outcomes in the sample space.

The numbers are $1, 2, 3, 4, 5$ and $6$.

Out of these, $2, 3$ and $5$ are prime numbers.

Thus the number of favourable outcomes is $3$.

So the probability would be:

$P$ = $\frac{(\text{number of favourable outcomes})}{(\text{total number of possible outcomes})}$

$P$ = $\frac{3}{6}$ = $\frac{1}{2}$ $\leftarrow\ Answer!$

From a bag containing $5$ red balls, $3$ green balls and $2$ white balls, two balls are drawn at random. Find the probability that the second ball is red given that the first ball drawn was already red.

First let us define the events.

$A$ = Event that the first ball is red.

$B$ = Event that the second ball is red.

We are interested in finding the probability $P(B|A)$.

For that we need $P(A$ and $B)$ and $P(A)$. So let us find them first.

$P(A)$ = $\frac{(\text{number of red balls})}{(\text{total number of balls})}$

$P(A)$ = $\frac{5}{10}$ = $\frac{1}{2}$

Event $(A$ and $B)$ = Two balls drawn at a time.

$2$ out of $5$ red balls can be chosen in $5C2$ = $10$ ways.

$2$ out of total $10$ balls can be chosen in $10C2$ = $45$ ways.

So the probability:

$P(A$ and $B)$ = $\frac{10}{45}$ = $\frac{2}{9}$

Thus the conditional probability:

$P(B│A)$ = $\frac{P(A\ and\ B)}{P(A)}$

$P(B│A)$ = $\frac{\frac{2}{9}}{\frac{1}{2}}$

$P(B│A)$ = $\frac{4}{9}$ $\leftarrow\ Answer$