#### definition

If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given by

P(E|F) = `(P(E ∩ F))/(P(F))` provided P(F) ≠ 0

#### notes

Consider the experiment of tossing three fair coins. The sample space of the experiment is

S = {HHH ,HHT,HTH , THH , HTT , THT , TTH , TTT }

Since the coins are fair, we can assign the probability `1/8` to each sample point. Let E be the event ‘at least two heads appear’ and

F be the event ‘first coin shows tail’. Then

E = {HHH, HHT, HTH, THH} and

F = {THH, THT, TTH, TTT}

Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})

= `1/8 + 1/8 + 1/8 + 1/8 = 1/2 `

and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})

= `1/8 + 1/8 + 1/8 + 1/8 = 1/2`

Also E ∩ F = {THH}

with P(E ∩ F) = P({THH}) = `1/8`

Now, the sample point of F which is favourable to event E is THH. Thus, Probability of E considering F as the sample space = `1/4`

or Probability of E given that the event F has occurred = `1/4`

This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F).

Thus P (E | F) = `1/4`

Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩ F.

Thus, we can also write the conditional probability of E given that F has occurred as

P (E|F) =

`("Number of elementary events favourable to E ∩ F")/("Number of elementary events which are favourable to F")`

`= (n(E ∩ F))/(n(F))`

Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that P(E|F) can also be written as

P(E|F) =` (n(E ∩ F))/(n(S)) / (n(F))/(n(S)) = (P(E ∩ F))/(P(F))` ...(1)

Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ

#### Video Tutorials

#### Shaalaa.com | Probability part 7 (Example :- Conditional Probability)

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