Answer :

poojan
probability of getting  the multiple of 2: (2×50=100)
no.of favorable outcomes=50
total no.of outcomes=100
let the event be P(E)
then P(E) = no.of favorable outcomes/total no.of outcomes
               = 50/100
               = 1/2
probability of getting  the multiple of 3: (3×33=99)
no.of favorable outcomes=33
total no.of possible outcomes=100
let the event be P(F)
then P(F) = 33/100
zaara786
n(s)=100
let 'A'be the event of getting a multiple of 2 and 3
A=(2,3,6,8,9,10,12,14,15,16,18,20,21,22,24,26,27,28,30,32,33,34,36,38,39,40,41,42,44,45,46,48,50,51,52,54,56,57,58,60,62,63,64,66,68,69,70,71,72,74,75,76,78,80,81,82,84,86,87,88,90,92,93,94,96,98,99,100)
n(s)=68
.:p(A)=n(a)/n(s)
.:      =68/100
.:      =17/25
ans=17/25

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