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Probability

Theoretical Probability and Experimental Probability

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.

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 …

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