The probability is the quantitative representation of the likelihood of an event to occur. In terms of probability, an event is referred to a possible set of outcomes when an experiment is performed and the probability is measured for the events. The probability of an event falls between $0$ and $1$. Here, $0$ represents an impossible event and $1$ implies a certain event. The chances of occurring of an event are more if the probability is higher.

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Numerically, the probabilities can be calculated by dividing the number of favorable or desired outcomes with the total number of possible outcomes. This is applicable in the cases when the experiments performed are well defined and random.

Let us take the simplest example. When an unbiased or fair coin is tossed in air, there are only two possible outcomes - head and tail. So, the probability to getting a head is equal to the $\frac{1}{2}$, where 1 is the favorable outcome (head) and $2$ represents total number of outcomes (head and tail). The coin being fair, the probability of both outcomes (head and tail) is same.

$P(E)$ = $\frac{Number\ of\ desired\ outcomes}{Number\ of\ total\ outcomes}$

The events that have only two possible outcomes are called complementary events.

**Example:** Tossing a coin $(\frac{head}{tail})$, getting $2$ and not $2$ on rolling a die.

$P(A) + P(A^C)$ = $1$

The probability of happening of either event A or event B is:

$P(A$ or $B)$ = $P(A) + P(B) - P(A$ and $B)$

We may even write this formula as:

$P(A \cup B)$ = $P(A) + P(B) - P(A \cap B)$

The events that cannot happen together at the same time are mutually exclusive. Two events $A$ and $B$ are mutually exclusive if

$P(A \cap B)$ = $0$

So, law of addition in that case is:

$P(A \cup B)$ = $P(A) + P(B)$

Two events are independent if happening of one doesnâ€™t affect that of another. So, events $A$ and $B$ are independent if

$P(A \cap B)$ = $P(A) \cdot P(B)$

Conditional probability refers to the probability of an event when another event has already occurred. Probability of event B, provided that A has already occurred is:

$P(A|B)$ = $\frac{P(A\ \cap\ B)}{P(B)}$

Bayes Law:

$P(A|B)$ = $\frac{[P(B|A)\ \cdot\ P(A)]}{P(B)}$

On a roll of a fair $6$ sided die, what is the probability of getting of rolling a $1$ or a $4$?

Total number of outcomes = $6$

Probability of getting a $1$

$P(1)$ = $\frac{1}{6}$

Probability of getting a $4$

$P(4)$ = $\frac{1}{6}$

Since both events are mutually-exclusive,

$P(1$ or $4)$ = $P(1) + P(4)$ = $\frac{1}{6}$ + $\frac{1}{6}$ = $\frac{1}{3}$

Determine the probability of drawing a jack and an ace consecutively without replacement from a well-shuffled deck of cards.

Total outcomes = $52$

There are $4$ jacks. So

$P(Jack)$ = $\frac{4}{52}$ = $\frac{1}{13}$

Since another card is drawn without replacement, hence now total outcomes are $51$.

$P(Ace)$ = $\frac{4}{51}$

Required probability = $P(Jack) \times P(Ace)$ = $\frac{1}{13}$ $\times$ $\frac{4}{51}$ = $\frac{4}{663}$