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* be the number of distinct sample points in the sample space S. Suppose, out of these *n* points, *m* points are favorable for the occurrence of event A. Hence, the remaining $n-m$ points are unfavorable for the occurrence of event A or we can say, $n-m$ points are favorable for the occurrence of event A'.

$\therefore P\left(A\right)=\frac{m}{n},P\left(A\text{'}\right)=\frac{n-m}{n}$

The ratio of favorable cases to the number of unfavorable cases is known as odds in the favor of event A which is given by $\frac{m}{n-m}$ i.e. P(A) : P(A').

The ratio of unfavorable cases to the number of favorable cases is known as odds against the favor of event A which is given by $\frac{n-m}{m}$ i.e. P(A') : P(A).

**Example 1:**

**One card is drawn from a well shuffled deck. What is the probability that the card will be**

**(i) a king?**

**(ii) not a king?**

**Solution:**

Let *E* be the event ‘the card is a king’ and *F* be the event ‘the card is not a king’.

(i) Since there are 4 kings in a deck.

Number of outcomes favourable to *E* = 4

Number of possible outcomes = 52

*P* (*E*)

- Here, the events E and F are complements of each other.

∴ *P(E) *+ *P(F)* = 1

*P(F)* = 1 −

**Example 2: **

**If the probability of an event A is 0.12 and B is 0.88 and they belong to the same set of observations, then show that A and B are complementary events.**

**Solution:**

It is given that *P *(*A*) = 0.12 and *P *(*B*) = 0.88

Now, P(A) + P(B) = 0.12 + 0.88 = 1

The events A and B are complementary events.

**Example 3:**

**Savita and Babita are playing badminton. The probability of Savita winning the match is 0.52. What is the probability of Babita winning the match?**

**Solution:**

Let *E* be the event ‘Savita winning the match’ and *F* be the event ‘Babita wining the match’.

It is given that *P *(*E*) = 0.52

Here, *E* and *F* are complementary events because if Babita wins the match, Savita will surely lose the match and vice versa.

*P *(*E*) + *P *(*F*) = 1

0.52 + *P *(*F*) = 1

*P *(*F*) = 1 − 0.52 = 0.48

Thus, the probability of Babita winning the match is 0.48.

**Example 4:**

**In a box, there are 2 red, 5 blue, and 7 black marbles. One marble is drawn from the box at random. What is the probability that the marble drawn will be (i) red (ii) blue (iii) black (iv) not blue?**

**Solution:**

Since the marble is drawn at random, all the marbles are equally likely to be drawn.

Total number of marbles = 2 + 5 + 7 = 14

Let *A* be the event ‘the marble is red’, *B* be the event ‘the marble is blue’ and *C* be the event ‘the marble is black.

**(i)** Number of outcomes favourable to event *A *= 2

*P *(*A*)

**(ii)** Number of outcomes favourable to event* B* = 5

∴ *P *(*B*)

**(iii)** Number of outcomes favourable to event *C* = 7

*P *(*C*)

**(iv)** We have, *P *(*B*)

The event of drawing a marble which is not blu…

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