Probability 8

When 6 is the lowest number, then the 3 dices are 6, 6 and 6.
6+6+6 < 6*5 => not possible
When 5 is the lowest number, and the sum 5+x+y >= 25 is also not possible.
When 4 is the lowest number, the sum 4+x+y >= 20 is also not possible.
When 3 is the lowest number, the sum 3+x+y>=15 is possible only when x=y=6 => 3c1 = 3 cases (366, 636, 663)
When 2 is the lowest number, the required minimum sum is 10. The following cases are less than 10:
2,2,2 => 1 case
2,2,3 => 3c1 = 3 cases
2,2,4 => 3c1 = 3 cases
2,2,5 => 3c1 = 3 cases
2,3,3 => 3c2 = 3 cases
2,3,4 => 3! = 6 cases
Total cases = 19
Total number of cases that all 3 dices does not contain 1 is (6-1)^3
Total number of cases that all 3 dices does not contain 1 and 2 is (6-2)^3
Total number of cases that all 3 dices does not contain 1 but contain 2 is 5^3 - 4^3
When 2 is the lowest number, cases for sum>=10 is
5^3 - 4^3 - 19
When 1 is the lowest number, the required minimum sum is 5. The following cases are less than 5:
1,1,1 => 1 case
1,1,2 => 3c1 = 3 cases
Total cases = 4
Total number of cases that at least one dice is 1 is 6^3 - 5^3
When 1 is the lowest number, cases for sum>=5 is
6^3 - 5^3 - 4

All in all, total possible cases are: 1 + 5^3 - 4^3 - 16 + 6^3 - 5^3 - 4
= 3 - 64 - 19 + 216 - 4
= 132

Probability = 132/216 = 11/18

If a single throw of three dice be made, what isthe probability that the
total number shown is not less than five times the lowest number
shown on any one of the three dice.
Sol
設 三次投出點數為a,b,c且a>=b>=c
(1) a，b，c 三次點數相同 =>N=0
(2) a，b，c 三次點數有二次相同
(21) a=b，b>c
(211) c=1 =>N=4*3!/(2!1!)=12
(212) c=2 =>N=2*3!/(2!1!1!)=6
(213) c=3 =>N=0
(22) a>b，b=c
(221) c=1 =>N=4*3!/(2!1!1!)=12
(222) c=2=>N=1*3!/(2!1!1!)=3
(3) a，b，c 三次點數全不相同
(31) c=1 =>N=10*3!/(1!1!11)=60
(32) c=2 =>N=6*3!/(1!1!1!)=36
(33) c=3 =>N=1*3!/(1!1!1!)=3
12+6+12+3+60+36+3=132
P=132/6^3=11/18