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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'.

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)

1. 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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