A fair die is thrown. If E is the event that 'the number appearing is a multiple of 3' and F be the event that the number appearing is even'. Then choose the correct option given below:

$P(F)=\frac{1}{3}$

Given that

E is the event that 'the number appearing is a multiple of 3' and

(Possible cases. 3,6,9,12)

F be the event that the number appearing is even'

(Possible Case. 2,4,6,8,10,12)

Then E∪F is the event such the number appeared  should be even and multiple of 3. The total no.  possibilities of such events are (6,12)

 So, P (E∪F)  = (Total no. of favorable cases/ Exhaustive no. of cases)

                    = 2/6

        P (E∪F)  = 1/3