Given that three fair dice have been tossed and the total of their top faces is found to be divisible by 3, but not divisible by 9. what is the probability that all three of them have landed 4?

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

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.