A die is thrown two times.
Total number of possible outcomes=36.
Let E is the event that sum of the score appearing on the die is multiple of 4.
Possible outcomes in favor of E=(1,3),(3,1),(2,2),(2,6),(6,2),(3,5),(5,3),(4,4),(6,6).
Number of favorable outcomes=9.
P(E)=41.
Let F is the event that at least one 4 appear.
Possible outcomes in favor of F=(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4),(4,5),(5,4),(4,6),(6,4)
E∩F=(4,4)
⇒P(E∩F)=361.
We have to find conditional probability of occurrence of at least one 4 given that sum of scores is multiple of 4.
P(F/E)=P(E)P(E∩F)=41361=91.