[中五數學] 有關機會率- Probability

Prime number即係質數，2粒dice，最大擲出12點，入面的Prime number有:
2,3,5,7,11
擲出的結果有36個:
(1,1)(02) (1,2)(03) (1,3)(04) (1,4)(05) (1,5)(06) (1,6)(07)
(2,1)(03) (2,2)(04) (2,3)(05) (2,4)(06) (2,5)(07) (2,6)(08)
(3,1)(04) (3,2)(05) (3,3)(06) (3,4)(07) (3,5)(08) (3,6)(09)
(4,1)(05) (4,2)(06) (4,3)(07) (4,4)(08) (4,5)(09) (4,6)(10)
(5,1)(06) (5,2)(07) (5,3)(08) (5,4)(09) (5,5)(10) (5,6)(11)
(6,1)(07) (6,2)(08) (6,3)(09) (6,4)(10) (6,5)(11) (6,6)(12)
「02」的結果有1個，「03」有2個，「05」有4個，「07」有6個，「11」有2個。
P(2)+P(3)+P(5)+P(7)+P(11)
=(1/36)+(2/36)+(4/36)+(6/36)+(2/36)
=5/12

2008-03-23 00:37:48 補充：
4樓就算有F7數學根柢，但答案未約簡，失左個A分
1樓就數都數錯，佢自己數都數錯，列出的可能性得14個，唔係20個呢。當然，佢冇考慮就 double 1可以出「2」這質數的可能性。
3樓仲勁.....fire dice，粒骰仔著火了嗎??

prime number是質數, 即是除1和自己以外沒有可以將其整除的數字

firstly, the population of outcome of the sum of the two numbers of the two dice are (2,3,4,5,6,7,8,9,10,11,12)

there are (2,3,5,7,11) 5 numbers which are prime numbers

to get 2, there are 1 way: (1,1)
to get 3, there are 2 ways: (1,2)(2,1)
to get 5, there are 4 ways: (1,4)(2,3)(3,2)(4,1)
to get 7, there are 6 ways: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)
to get 11, there are 2 ways:(5,6)(6,5)

for each way, the probability is 1/6 X1/6=1/36

there are (1+2+4+6+2)=15 ways to get a prime number,

so, the probability of getting a prime number =15 X 1/36=15/36

prime number 是 質數。

e 家有 2 粒骰，佢要搵 2 個點數的和係質數的機會率。
方法如下：

1) 有機會出現的情況：

1 /2 /4 /6
2 /1 /3 /5
3 /2 /4
4 /1 /3
5 /2 /6
6 /1 /5
=> 出現質數的情況 有 20 個

2) 總共會出現的組合： 6x6 = 36

3) P( the sum of the two numbers thrown is a prime number) = 20/36
= 5/9