Probability

Sample Space and Events of Experiments

Consider the experiment of throwing a dice. Any of the numbers 1, 2, 3, 4, 5, or 6 can come up on the upper face of the dice. We can easily find the probability of getting a number 5 on the upper face of the dice?

Mathematically, probability of any event *E* can be defined as follows.

Here, *S* represents the sample space and *n*(*S*) represents the number of outcomes in the sample space.

For this experiment, we have

Sample space (*S*) = {1, 2, 3, 4, 5, 6}. Thus, *S* is a finite set.

So, we can say that the possible outcomes of this experiment are 1, 2, 3, 4, 5, and 6.

Number of all possible outcomes = 6

Number of favourable outcomes of getting the number 5 = 1

Probability (getting 5)

Similarly, we can find the probability of getting other numbers also.

*P* (getting 1), *P* (getting 2), *P *(getting 3), *P* (getting 4) and

*P* (getting 6)

Let us add the probability of each separate observation.

This will give us the sum of the probabilities of all possible outcomes.

*P* (getting 1) + *P *(getting 2) + *P* (getting 3) + *P* (getting 4) + *P* (getting 5) + *P* (getting 6) = +++++ = 1

“**Sum of the probabilities of all elementary events is 1”.**

Now, let us find the probability of **not** getting 5 on the upper face.

The outcomes favourable to this event are 1, 2, 3, 4, and 6.

Number of favourable outcomes = 5

P (not getting 5)

We can also see that *P *(getting 5) + *P *(not getting 5)

“**Sum of probabilities of occurrence and non occurrence of an event is 1”.**

i.e. **If E is the event, then **

**P****(**

**E****) +**

**P****(not**

**E****) = 1**… (1)

or we can write **P(E) = 1 ****− ****P (****not ****E****)**

Here, the events of getting a number 5 and not getting 5 are complements of each other as we cannot find an observation which is common to the two observations.

Thus, event **not *** E *is the complement of event

*E.*Complement of event

*E*is denoted by or

*E*

**'**.

Using equation (1), we can write

P (E) + P () = 1 |

or

P () = 1 – P (E) |

This is a very important property about the probability of complement of an event and it is stated as follows:

**If E is an event of finite sample space S, then **

**P****() = 1 –**

*P*(*E*) where is the complement of event*E*.Now, let us prove this property algebraically.

**Proof:**

We have,

*E* ∪ = *S* and *E* ∩ =

⇒ *n*(*E* ∪ ) = *n*(*S*) and *n*(*E* ∩ ) = *n*()

⇒ *n*(*E* ∪ ) = *n*(*S*) and *n*(*E* ∩ ) = 0 ...(1)

Now,

*n*(*E* ∪ ) = *n*(*S*)

⇒ *n*(*E*) + *n*() – *n*(*E* ∩ ) = *n*(*S*)

⇒ *n*(*E*) + *n*() – 0 = *n*(*S*) [Using (1)]

⇒ *n*() = *n*(*S*) – *n*(*E*)

On dividing both sides by *n*(*S*), we get

**⇒ P() = 1 – P(E)**

Hence proved.

Let us solve some examples based on this concept.

**ODDS (Ratio of two complementary probabilities):**

Let *n* …

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