An event is considered a subset of a sample space.

If we toss a die once the sample space is the set S = {1, 2, 3, 4, 5, 6} where each outcome is equally likely to happen.

If we want the event to be the number 3 on the die when it is tossed then the subset of the sample space is {3}.

It is a singleton set. Such an even this called a **simple event**.

If however we consider the event as 'getting a prime number' the subset of the sample space is {2, 3, 5} which has 3 elements.

Such an event is called a compound event.

If we toss a die, we can get an 'even number' or an 'odd number' but not both. So the event A of getting an even number excludes the event B of getting an odd number and vice versa. We call such events **mutually exclusive events**.

When we toss a coin, getting a head or getting a tail are mutually exclusive.

If we draw a card and our events are

A: getting a spade

B: getting a king

We can get the king of spades so these events are not mutually exclusive.

Suppose a bag contains 4 red balls and 8 black balls. Two balls are drawn one after the other without the first being replaced. Let A be the event of "getting a black ball in the first draw" and B be of "getting a red ball" in the 2nd draw. Suppose the event A occurs that is we get a black ball in the first draw then P (B) = 4/11 (4 red balls out of 12 – 1 = 11 balls).

If however A does not occur that is, instead of drawing a black ball, a red ball is drawn first, then P (B) = 3/11 (4 - 1 red balls out of 12 –1 = 11 balls)

Obviously P(B) depends on whether A occurs or not.

In such a case the events are called **Dependent Events**.

On the other hand, if we toss a coin twice and event A is ‘getting a head in the first toss' event B is ‘getting a tail in the second toss’, P (B) remains the same whether we get a head in the first toss or not. Such events are called Independent Events.

We know that the probability of getting a number 6 in the throw of a dice is 1/6. What is the probability of getting a number other than 6? The numbers are 1, 2, 3, 4 or 5 or 5 favorable outcomes.

Sometimes we hear people say the 'Odds of a particular team winning are 3 to 1'.

Suppose we shuffle a deck of cards what are the odds that the first card we draw is an ace?

Since there are 4 aces we know that the probability of getting an ace is

Probability of not getting an ace

The probability of **not** drawing an ace is 12 times the probability of getting an ace. So the **odds** of getting an ace is 1 to 12 or the **odds** of not getting an ace is 12 to 1.

We can now form the following definition:

Suppose there are **m** outcomes favorable to an event and n outcomes which are unfavorable to the event in the sample space then

What are the odds in favor of getting a '4' in a throw of a die? What are the odds against getting a '4'

**Solution:**

Number of favorable outcomes = 1

Number of unfavorable outcomes = 6 - 1 = 5

We can say that when we express odds in terms of probability then

The odds in favor of an event are 4 to 5. Find the probability that it will occur

**Solution:**

Odds in favor of the event occurring is 4/5. Thus

Probability that it will occur = 4/9.

If P(A) = 3/5 and P(B) = 1/3 find

a. P (A or B) if A and B are mutually exclusive events

b. P(A and B) if A and B are independent events

**Answer:**

- P(A) = 3/5 P(B) = 1/3 A and B are mutually exclusive events P(A or B) = P(A ∪ B) = P(A) + P(B) by the addition theorem of probability
- A and B are independent events P(A and B) = P(A ∩ B) = P(A) * P(B) By the addition theorem of probability

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