Originally Posted by

**thereddevils** A fair tetrahedron die has four faces , each with digit 1,2,3,4 . THe die is tossed three times. The sum of three visible surfaces is recorded as T at each toss . The events A , B and C are defined as follows .

A= The score T at the first toss is 7 .

B=SUm of first 2 scores is less than 15

C=Product of the three scores is a multiple of 3 .

FInd P(A) , P(B) and P(C)

For P(A) , It can only (1,3,4) and this can be arranged in 3!=6 ways so

P(A)=6/24=1/4

For P(B) , its either (6,8) , (8,6) or (6,6)

6=(1,2,3) , 8=(1,3,4)

so p(B)=3 x (3!/4!)(3!/4!)=3/16 but the answer given is 3/8