When the occurrence of an event does not affect the probability occurrence of another event, then we say that the events are independent.

If the occurrence of an event affects the probability of occurrence of another event then the events are said to be dependent.
The probability of occurrence of two events together is the probability of intersection of the two events. It is denoted by $P (M \cap N)$. In cases where we have mutually exclusive events then $P (M \cap N)$ = 0.

Rules

The multiplication rule in probability is applicable on two events in cases where the events are occurring as an output of more than one experiment.

The rule states that: “the probability of occurrence of two events that is the probability of intersection of two events is same as the product of probability of occurrence on one event and the probability of occurrence of the second event.”

That is, if $X$ and $Y$ are two events then:
$P (X and Y)$ = $P (X) * P (Y)$ or $P (X \cap Y)$ = $P (X) * P (Y)$

This is valid in all cases whether the events are independent or dependent. Indeed in case of dependent events one has to be careful while finding the probability of second event.

In that case we can also modify the multiplication rule as:

$P (X and Y)$ = $P (X) * P (Y|X)$ or

$P (X \cap Y)$ = $P (X) * P (Y|X)$

Here, $P (Y|X)$ implies the probability of occurrence of event Y when event X has already taken place.

For Example: consider the events of drawing two balls from a box of marble balls one by one with replacement. In this case the events are independent and thus direct product of probabilities of the two events can be taken in order to obtain the probability of both events together. 

In other case, suppose we are taking out two balls from the box of marble balls without replacing the first one. In this case, the events are dependent and so we will be required to find the conditional probability in order to find the probability of occurrence of both events together.

Examples

Let us take up a few examples on multiplication rule of probability.

Example 1: A basket contains 3 oranges and 7 apples. Two fruits are taken with replacement fm the basket. Find the probability that both of the fruits are apples.

Solution: Let $X$ = event that first fruit is apple and $Y$ = event that second fruit is apple.
Since the fruits are replaced back after each pick, this implies that the events are mutually exclusive events.

So by multiplication rule we have,

$P (X \cap Y)$ = $P (X) * P (Y)$

P (X) = $\frac{7}{10}$ and P (Y) = $\frac{7}{10}$

$\rightarrow$ $P (X \cap Y)$ = $\frac{7}{10}$ * $\frac{7}{10}$ = $\frac{49}{100}$ = 0.49

Example 2: We have 10 red and 5 blue balls in a bag. We draw a ball from the bag and set it aside. We draw another ball from the bag now. Find the probability that both balls are blue.

Solution: Let $X$ = event that first ball is blue and $Y$ = event that second ball is blue.
From question, $P (X)$ = $\frac{5}{15}$.

Now $P (Y)$ = $P (Y|X)$ [since the events are dependent]

$\rightarrow$ $P (Y)$ = $\frac{4}{14}$

So, $P (X \cap Y)$ = $\frac{5}{15}$ * $\frac{4}{14}$ = $\frac{20}{210}$ = $\frac{2}{21}$