Probability is classified broadly in two ways: theoretical and experimental. In theoretical probability we take the likelihood of occurrence of an event whereas experimental probability is based on the occurrence of an event while an experiment is conducted or a data is collected.

Theoretical probability, $P(E)$ = $\frac{n(E)}{n(S)}$

where, S is the sample space and E is the event.

Find the theoretical probability of getting a head when a coin is tossed.

Sample space, $n(S) = 2$

Number of likely events where head will come, $n(E) = 1$

Probability of getting head, $P(E)$ = $\frac{1}{2}$

A coin is tossed twice, then what will be be the probability of getting one head and one tail?

Sample space, $S = {HH, HT, TH, TT}$

$n(S) = 4$

Events getting a head and a tail, $E = {HT, TH}$

$n(E) = 2$

$P(E)$ = $\frac{n(E)}{n(S)}$ = $\frac{2}{4}$ = $\frac{1}{2}$

Hence, the probability of getting one head and one tail is $\frac{1}{2}$

A bag has $3$ red, $2$ green and $4$ black ball. If a ball is randomly taken from the bag, find the theoretical probability that it is green.

Solution:

Sample space, $n(S) = 3 + 2 + 4 = 9$

$n(E)$ = number of green balls = $2$

Probability of getting a green ball, $P(E)$ = $\frac{2}{9}$

In a deck of $52$ cards, one card is chosen which is black. Another card is chosen without replacement then what is the probability that it is also a black card?

Solution:

Sample space, $n(S)$ = total number of cards = $51$ (as one card is already taken out)

Number of favorable events, $n(E)$ = number of black cards = $25$

Probability, $P(E)$ = $\frac{25}{51}$

If a dice is rolled $4$ times and $3$ times an even number comes then what is the theoretical probability of getting an even number?

Solution:

If a dice is rolled then the sample space, $S = {1, 2, 3, 4, 5, 6}$

$n(S) = 6$

$n(E) = {2, 4, 6} = 3$

Probability of getting an even number, $P(E)$ = $\frac{3}{6}$ = $\frac{1}{2}$

Find the theoretical probability of a bulb to be switched on.

Solution:

Sample space, $S = {ON, OFF}$

$n(S) = 2$

$n(E) = {ON} = 1$

$P(E)$ = $\frac{1}{2}$