中五數學-概率問題

If the 3 appear at the first dice
Then the possible combinations of the other two dices
(1,2) (1,4) (1,6) (2,1) (2,5) (4,1) (4,5) (5,2) (5,4) (5,6) (6,1) (6,5)
Similar for 3 appears at the second or third dice
so total combinations 36
probability
=36/216
=1/6
2
(a)
probability=(4/7)(1/3)(3/5)=4/35
(b)
probability
=(4/7)(2/3)(2/5)+(3/7)(1/3)(2/5)+(3/7)(2/3)(3/5)
=(16+6+18)/105
=40/105
=8/21
(c)
the probability that the problem cannot solved
=(3/7)(2/3)(2/5)
=12/105
So the probability problem can be solved
=1-12/105
=93/105
(d)
in this case the probability problem cannot be solved
=(1/3)(3/7+2/3+2/5)
=[(45+70+42)/105](1/3)
=157/315
So probability that the problem can be solved
=158/315
Right?

problem is solved independently by Alfred, Sam,Andrew.The probability that the problem can be solved by them are "4/7","1/3","3/5".
find the probability that