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Conditional Probability

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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


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

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 

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