Answer:
Probability = 1/45
Step-by-step explanation:
Since the numbers from 3 to 18 that are divisible by 3 but not by 9 include
(3,6,12,15)
Now the case in our favor is that all die have landed 4 thus the sum is 12 and hence a valid condition
for case with sum as 3 we have the following individual sub cases
(1,1,1) i.e each die lands 1
For the case of sum as 6 we have the following cases
(1,1,4) and it's permutations=(1,1,4),(1,4,1),(4,1,1)=3
(2,2,2) =1
(1,2,3) and it's permutations=3! = 6
Total cases for sum as 6 becomes 10
For the case of sum equals 12 we have the following cases of permutation
(6,5,1) in 3! =6 ways
(6,4,2) in 3!=6 ways
(6,3,3) in 3!/2!=3 ways
(5,5,2) in 3!/2!=3 ways
(5,4,3) in 3! =6 ways
(4,4,4) in 1 way
total of 6+6+3+3+6+1=25 ways
For the case of sum equals 15 we have the following cases of permutation
(6,6,3) in 3!/2!=3 ways
(6,5,4) in 3! =6 ways
total of 6+3=9 ways
Thus total possible outcomes are 1+10+25+9 = 45 ways
Thus probability is given by (Number of favorable cases)/(Total Cases)
Thus P(E)= 1/45
