# Theoretical Probability orClassical Probability

Moving forward to the theoretical probability which is also known as classical probability or priori probability, we will first discuss about collecting all possible outcomes and equally likely outcome.

Collecting all Possible Outcomes:

When an experiment is done at random we can collect all possible outcomes without actually doing the experiment repeatedly.

For example:

1. If a coin is tossed, either a head (H) or a tail (T) will show.
2. If a die is rolled, it will show either 1 or 2 or 3 or 4 or 5 or 6.
3. If two coins are tossed simultaneously, either HH or HT or TH or TT will show. (TH means tail on the first coin and head on the second coin.)

Thus, the collection of all possible outcomes in tossing a coin consists H, T. So, there are only two different outcomes in tossing a coin.

The collection of all possible outcomes in throwing a die consists of 1, 20, 3, 4, 5, 6. So, there are only six different outcomes in a trail of throwing a die.

The collection of all possible outcomes in tossing two coins simultaneously consists of HH, HT, TH, TT. So, there are only four different outcomes in a trail of tossing two coins.

Equally Likely outcome:

When an experiment is done at random, any one of the possible outcomes may take place. If the possibility of each outcome taking place is the same, we say the outcomes are equally likely.

If a perfectly manufactured coin is tossed, the outcome H (head) and the outcome T(tail) are equally likely. But if half of the coin on head's side is heavier then it is more likely that T will appear on the top. So, if a defective (biased) coin is tossed the outcomes H and T are not equally likely. In what follows all the outcomes in a trail will be assumed to be equally likely.

Classical Probability: The classical probability of an event E, denoted by P (E) is defined as below

P(E) = $$\frac{\textrm{Number of Outcomes Favourable to the Event E}}{\textrm{Total Number of Possible Outcomes in the Experiment}}$$

Definition of Theoretical Probability:

Let a random experiment produce only finite number of mutually exclusive and equally likely outcomes. Then the probability of an event E is defined as

Number of favorable outcomes
P(E) =     Total number of possible outcome

The formula for finding the theoretical probability of an event is

Number of favorable outcomes
P(E) =     Total number of possible outcome

Theoretical probability is also known as Classical or A Priori probability.

To find the theoretical probability of an event we need to follow the above explanation.

Problems based on Theoretical Probability or Classical Probability:

1. A fair coin is tossed 450 times and the outcomes were noted as: Head = 250, Tail = 200.

Find the probability of the coin showing up

(ii) a tail.

Solution:

Number of times coin is tossed = 450

Number of tails = 200

(i) Probability of getting a head

Number of favorable outcomes
P(H) =     Total number of possible outcome

= 250/450
= 5/9

(ii) Probability of getting a tail

Number of favorable outcomes
P(T) =     Total number of possible outcome

= 200/450
= 4/9

2. In a cricket match the Sachin hit a boundary 5 times out of 30 balls he plays. Find the probability that he

(i) hit a boundary

(ii) do not hit a boundary.

Solution:

Total number of balls Sachin played = 30

Number of boundary hit = 5

Number of times he did not hit a boundary = 30 - 5 = 25

(i) Probability that he hit a boundary

Number of favorable outcomes
P(A) =     Total number of possible outcome

= 5/30
=1/6

(ii) Probability that he did not hit a boundary

Number of favorable outcomes
P(B) =     Total number of possible outcome

= 25/30
= 5/6

3. The record of weather stations report shows that out of the past 95 consecutive days, its weather forecast was correct 65 times. Find the probability that on a given day:

(i) it was correct

(ii) it was not correct.

Solution:

Total number of days = 95

Number of correct weather forecast = 65

Number of not correct weather forecast = 95 - 65 = 30

(i) Probability of ‘it was correct forecast’

Number of favorable outcomes
P(X) =     Total number of possible outcome

= 65/95
= 13/19

(ii) Probability of ‘it was not correct forecast’

Number of favorable outcomes
P(Y) =     Total number of possible outcome

= 30/95
= 6/19

4. In a society 1000 families with 2 children were selected and the following data was recorded

Find the probability of a family, having:

(i) 1 boy

(ii) 2 boys

(iii) no boy.

Solution:

According to the given table;

Total number of families = 333 + 392 + 275 = 1000

Number of families having 0 boy = 333

Number of families having 1 boy = 392

Number of families having 2 boys = 275

(i) Probability of having ‘1 boy’

Number of favorable outcomes
P(X) =     Total number of possible outcome

= 392/1000
= 49/125

(ii) Probability of having ‘2 boys’

Number of favorable outcomes
P(Y) =     Total number of possible outcome

= 275/1000
= 11/40

(iii) Probability of having ‘no boy’

Number of favorable outcomes
P(Z) =     Total number of possible outcome

= 333/1000

More solved examples on theoretical probability or classical probability:

5. Two fair coins are tossed 225 times simultaneously and their outcomes are noted as:

(i) Two tails = 65,

(ii) One tail = 110 and

(iii) No tail = 50

Find the probability of occurrence of each of these events.

Solution:

Total number of times two fair coins are tossed = 225

Number of times two tails occur = 65

Number of times one tail occur = 110

Number of times no tail occur = 50

(i) Probability of occurrence of ‘two tails’

Number of favorable outcomes
P(X) =     Total number of possible outcome

= 65/225
= 13/45

(ii) Probability of occurrence of ‘one tail’

Number of favorable outcomes
P(Y) =     Total number of possible outcome

= 110/225
= 22/45

(iii) Probability of occurrence of ‘no tail’

Number of favorable outcomes
P(Z) =     Total number of possible outcome

= 50/225
= 2/9

6. A die is thrown randomly four hundred fifty times. The frequencies of outcomes 1, 2, 3, 4, 5 and 6 were noted as given in the following table:

Find the probability of the occurrence of the event

(i) 4

(ii) a number < 4

(iii) a number > 4

(iv) a prime number

(v) a number < 7

(vi) a number > 6

Solution:

Total number of times a die is thrown randomly = 450

(i) Number of occurrence of a number 4 = 75

Probability of the occurrence of ‘4’

Number of favorable outcomes
P(A) =     Total number of possible outcome

= 75/450
= 1/6

(ii) Number of occurrence of a number less than 4 = 73 + 70 + 74 = 217

Probability of the occurrence of ‘a number < 4’

Number of favorable outcomes
P(B) =     Total number of possible outcome

= 217/450

(iii) Number of occurrence of a number greater than 4 = 80 + 78 = 158

Probability of the occurrence of ‘a number > 4’

Number of favorable outcomes
P(C) =     Total number of possible outcome

= 158/450
= 79/225

(iv) Number of occurrence of a prime number i.e. 2, 3, 5 = 70 + 74 + 80 = 224

Probability of the occurrence of ‘a prime number’

Number of favorable outcomes
P(D) =     Total number of possible outcome

= 224/450
= 112/225

(v) Number of occurrence of a number less than 7 i.e. 1, 2, 3, 4, 5 and 6 = 73 + 70 + 74 + 75 + 80 + 78 = 450

Probability of the occurrence of ‘a number < 7’

Number of favorable outcomes
P(E) =     Total number of possible outcome

= 450/450
= 1

(vi) Number of occurrence of a number greater than 6 = 0,

Because when a die is thrown all the 6 outcomes are 1, 2, 3, 4, 5 and 6

so, there is no number greater than 6.

Probability of the occurrence of ‘a number > 6’

Number of favorable outcomes
P(F) =     Total number of possible outcome

= 0/450
= 0

Solved example problem on classical probability:

7. Find the probability of getting a composite number in a throw of a die.

Solution:

Let E = the event of getting a composite number.

Total number of possible outcomes = 6 (Since any one of 1, 2, 3, 4, 5, 6 can come).

Number of favourable outcomes for the event E = 2 (Since any one of 4, 6 is a composite number).

Therefore,

P(E) = $$\frac{\textrm{Number of Outcomes Favourable to the Event E}}{\textrm{Total Number of Possible Outcomes}}$$

= $$\frac{2}{6}$$

= $$\frac{1}{3}$$.

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