Probability is a prediction that an event will occur. Probability of an event lies between zero and one. There are two types of events commonly when we talk about probability: mutually exclusive and mutually inclusive.

Mutually exclusive events are the ones when they cannot happen at same time, that is, there is no outcomes common in these events. For example: getting a head or a tail when a coin is tossed. On tossing a coin, we can never get head or tail together, so these two events are mutually exclusive.

Mutually exclusive events are the ones when they cannot happen at same time, that is, there is no outcomes common in these events. For example: getting a head or a tail when a coin is tossed. On tossing a coin, we can never get head or tail together, so these two events are mutually exclusive.

In this case the combined probability of two events can be obtained by simply adding up the individual properties of the events:

P $(X \cup Y)$ = P (X) + P (Y), where X and Y are mutually exclusive events.

Mutually inclusive events are the ones in which there are some common outcomes in between the given events. Like getting an odd number or getting a prime number when we throw a dice. In these two events there are common outcomes {3, 5} repeating in both the events. So these two events are mutually inclusive events.

In case of mutually inclusive events, we can evaluate the combined probability of the two events with the help of following formula:

P $(M \cup N)$ = P (M) + P (N) - P $(M \cap N)$

Where, P $(M \cup N)$ is the combined probability of occurrence of either M or N,

P (M) is the probability of occurrence of event M, P (N) is the probability of occurrence of event N, and P $(M \cap N)$ is the probability of joint occurrence of both M & N.

When there is no common outcome between M and N then P $(M \cap N)$ = 0

Let us see some problems on mutually inclusive events for better knowledge.

Which of the following events are mutually inclusive and which are not? Give reasons.

Clearly n (S) = 6

Let A = number greater than 3 = {4, 5, 6}

=> P (A) = $\frac{3}{6}$

=> P (A) = $\frac{3}{6}$

Let B = multiple of 2 = {2, 4, 6}

=> P (B) = $\frac{3}{6}$

=> P (B) = $\frac{3}{6}$

$A \cap B$ = {4, 6}

=> P (A $\cap$ B) = $\frac{4}{6}$

=> P (A $\cap$ B) = $\frac{4}{6}$

Hence the two events are mutually inclusive.

Hence P (A $\cup$ B) = P (A) + P (B) - P (A $\cap$ B) = $\frac{3}{6}$ + $\frac{3}{6}$ - $\frac{2}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$

Hence P (A $\cup$ B) = P (A) + P (B) - P (A $\cap$ B) = $\frac{3}{6}$ + $\frac{3}{6}$ - $\frac{2}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$