Answer:
Probability of getting sum less than
10 = Â \frac{30}{36}= Â \frac{5}{6}
Probability of getting a multiple of 3= \frac{20}{36}= \frac{5}{9} Â
(A) P(B|A)= \frac{P(B∩A)}{P(A)} = \frac{15/36}{30/36}= \frac{15}{30}=0.5 Â
(B)P(A|B)= \frac{P(A∩B)}{P(B)}= \frac{15/36}{5/9}=0.75 Â
(C) {A∩B} = {3, 6, 9, 12, 15, 18}
(D) {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}
-Hope this helps-