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Non Mutually Exclusive Events

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In probability theory, events are broadly classified into mutually exclusive and non-mutually exclusive events. If two or more events are such that they cannot occur simultaneously, then they are said to be mutually exclusive events. For example, the events that ‘it will rain’ and ‘it will not rain’ are mutually exclusive events. It is not possible that it will rain and it will not rain at the same time. However, if two events are like this: ‘it will rain’ and ‘waterlogging will happen’. Both these events are possible simultaneously. Thus these events are non-mutually exclusive events. 

The same concept is true for more than two events as well. We can have three, four or more events are may be mutually exclusive or mutually non-exclusive events.

Addition Theorem based on Mutually Non-Exclusive Events

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First, let us take a look at the addition theorem for mutually exclusive events. Consider the experiment of rolling a fair dice. Let us define two events such that, 

Event A = the number thrown up is $2$ or $3$

Event B = the number thrown up is $4$, $5$ or $6$.

Now, since there are 6 possible outcomes in the sample space, the probabilities of both the above events would be:

$P(A)$ = $\frac{2}{6}$ = $\frac{1}{3}$

And

$P(B)$ = $\frac{3}{6}$ = $\frac{1}{2}$

Now both these events are mutually exclusive, since, both cannot occur simultaneously. Therefore the combined probability of both these events would be:

$P(A\ or\ B)$ = $\frac{1}{3}$ + $\frac{1}{2}$ = $\frac{2\ +\ 3}{6}$ = $\frac{5}{6}$

We see that effectively the probability of occurrence of event A or event B is the sum of the individual probabilities of event A and event B. Thus for mutually exclusive events, 

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

Now for the same experiment, consider the following events:

Event C = the number thrown up is even.

Event D = the number thrown up is prime.

The number of ways in which event C can occur is $3$, since there are three even numbers $2$, $4$ and $6$. The number of ways in which event D can occur is also $3$, since there are $3$ prime numbers: $2$, $3$ and $5$. Thus, the probability of both the events would be:

$P(C)$ = $\frac{3}{6}$ = $\frac{1}{2}$

And

$P(D)$ = $\frac{3}{6}$ = $\frac{1}{2}$

Now, this time, the events are not mutually exclusive. As we can see that both the events conditions are fulfilled if the number $2$ throws up, since $2$ is even as well as prime. Thus the event C and D is not a null set. So,

$P(C\ \cap\ D)$ = $\frac{\text{number of favourable events}}{\text{total number of possible events}}$ = $\frac{1}{2}$

So now in this case the set of events $C$ or $D$ would be:

$C\ \cap\ D$ = {$2$, $3$, $4$, $5$, $6$}

Thus, number of elements in this set is:

$n(C\ \cap\ D)$ = $5$

Therefore, 

$P(C\ \cup\ D)$ = $\frac{5}{6}$

Thus we see that 

$P(C\ \cup\ D)\ \neq P(C)\ +\ P(D)$

This brings us to the formula for the addition theorem for mutually non-exclusive events
It is like this:

$P(C\ \cup\ D)$ = $P(C)\ +\ P(D)\ -\ P(C\ \cap\ D)$
Let us plug in the values for the above and check for our example:

$\frac{5}{6}$ = $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{6}$

$\frac{5}{6}$ = 1 - $\frac{1}{6}$

$\frac{5}{6}$ = $\frac{6\ -\ 1}{6}$

$\frac{5}{6}$ = $\frac{5}{6}$

$LHS$ = $RHS$

Thus we see that it holds!

Applications of Non-Mutually Exclusive Events

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The concept of non-mutually exclusive events finds lots of applications in the real world. Consider a situation wherein students of a class are asked which sport they like: Football, baseball or basketball. It is possible that there are a few students who like two of the three or all three. Thus the events would be mutually non-exclusive. 

Another example: A college offers courses for studying three different subjects, Math, English, and Arts. Some students may opt for one of these courses, some students can opt for two of these courses but there will be some students who opt for all the three. So these events are also mutually non-exclusive.

Three Non-Mutually Exclusive Events

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In the two applications that we saw above, note that there are three events that are not mutually exclusive. So far we have only seen the addition formula for $2$ events that are not mutually exclusive. Let us now understand how to calculate combine probability for three such events. The combined probability would be:

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

The formula is very similar to the set theory formula for a cardinal number of the union of three sets. Let us work out a few examples based on this formula to understand it better.

Examples

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Example 1: A dice is rolled. The following events are described:

Event A = even number is thrown up.

Event B = prime number is thrown up.

Event C = digits 2 or 3 are thrown up. Find the probability of $A\ \cup\ B\ \cup\ C$.

Solution:

The sample space consists of the following:

$1,\ 2,\ 3,\ 4,\ 5,\ 6$

So there are $6$ possible outcomes in the sample space.

For event $A$ there are $3$ possible outcomes as there are $3$ even numbers.

Thus, 

$P(A)$ = $\frac{3}{6}$ = $\frac{1}{2}$

For event B there are again 3 possible outcomes. So the probability would be:

$P(B)$ = $\frac{3}{6}$ = $\frac{1}{2}$

For event $C$, there are $2$ outcomes. Thus probability:

$P(C)$ = $\frac{2}{6}$ = $\frac{1}{3}$

$P(A\ \cap\ B)$ = $\frac{1}{6}$

$P(B\ \cap\ C)$ = $\frac{2}{6}$ = $\frac{1}{3}$

$P(C\ \cap\ A)$ = $\frac{1}{6}$

$P(A\ \cap\ B\cap\ C)$ = $\frac{1}{6}$


Thus plugging these into the formula we have:

$P(A\ \cup\ B\ \cup\ C)$ = $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{6}$ - $\frac{1}{3}$ - $\frac{1}{6}$ + $\frac{1}{6}$ = $\frac{5}{6}$  $\leftarrow$ Answer
Examples 2: A card is drawn from a standard deck of $52$ cards. Find the probability of drawing a spade or an ace.

Solution:

Number of spades = $13$

Therefore probability of spades:

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

Number of aces = $4$

Therefore probability of aces:

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

Number of spade aces = $1$

Therefore probability of spade ace:

P(spade & ace) = $\frac{1}{52}$

Probability of spade or ace would be:

$P(spade\ or\ ace)$ = $P(spade)$ + $P(ace)$ - P(spade & ace)

$P(S\ \cup\ A)$ = $\frac{1}{4}$ + $\frac{1}{13}$ - $\frac{1}{52}$ = $\frac{13\ +\ 4\ -\ 1}{52}$ = $\frac{16}{52}$ = $\frac{4}{13}$  $\leftarrow$ Answer