A** compound event** is one in which there is more than one possible outcome.

Tossing a die is a simple event. Tossing two dice is a compound event. The probability of a compound event can be calculated if its outcomes are equally likely.

**Example showing difference between mutually exclusive and not mutually exclusive events:**

- A pair of dice is rolled. The events of rolling a 6 and of rolling a double have the outcome (3,3) in common. These two events are NOT mutually exclusive.
- A pair of dice is rolled. The events of rolling a 9 and of rolling a double have NO outcomes in common. These two events ARE mutually exclusive.

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There are two types of compound events:

- If two events, A and B are mutually exclusive, then the probability that either A or B occurs is the sum of their probabilities.

- It is a situation where one event cannot occur with the other.
- If Two events (say A and B) are inclusive, then the probability that either A or B occurs is the sum of their probabilities subtracting the probability of both events occurring.

- This may be done by using a tree diagram.
- By using formula explained further.

Above we have described two types of compound events. There are two different formulas to explain each as listed below:

**Say A and B are two events, then**

**For Mutually exclusive events :**

P(A or B) = P (A) + P(B)

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

The probability of choosing a king = $P(A)$ = $\frac{4}{52}$

And the probability of choosing a queen of heart = $P(B)$ = $\frac{1}{52}$

So, the compound probability is: $P(A\ or\ B)$ = $P(A)\ +\ P(B)$ = $\frac{4}{52}\ +\ \frac{1}{52}$ = $\frac{5}{52}$.

Are these events mutually exclusive or inclusive? What is the probability that a patient has either high blood pressure or arteriosclerosis?

These are mutually inclusive events (Why?) since both can occur at the same time.

The probability of having high blood pressure = $P(A)$ = $\frac{3}{8}$

The probability of having arteriosclerosis = $P(B)$ = $\frac{5}{12}$

The probability of having both = $P(A\ and\ B)$ = $\frac{1}{4}$.

Using formula : $P\ (A\ or\ B)$ = $P(A)\ +\ P(B)\ –\ P(A\ and\ B)$

We get, $\ P\ (A\ or\ B)$ = $\frac{3}{8}\ +\ \frac{5}{12}\ –\ \frac{1}{4}$ = $\frac{13}{24}$

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