簡單概率一問~

1. 3/5 X 2/5 = 6/25

2a.能被５整除的奇數只有５，１５，２５．．．
10/100=1/10

2b.能被１０整除的４的倍數只能２０，４０，６０．．
5/100=1/20

Pay very special attention to Question 2 !!!


1) P(pick 1 boy and 1 girl)

= P(pick boy from Mr. Wong and pick girl from Mr. Lee) + P(pick girl from Mr. Wong and pick boy from Mr. Lee)

= 3/5 X 3/5 + 2/5 X 2/5

= 13/25



For question 2, it is conditional probability!!! Both previous answer has committed a very serious error.

Please note that:

(a) P(該數能被５整除，已知該數是奇數) is NOT equal to P(該數能被５整除 AND 該數是奇數)

(b) P(該數能被10整除，已知該數是4的倍數) is NOT equal to P(該數能被10整除 AND 該數是4的倍數)



(a)

~ 已知該數是奇數: so possible outcome are 1, 3, 5, 7... there is only 50 possible outcome (so 分母 is 50, NOT 100).

~ 該數能被５整除: WITHIN those 50 choices, possible outcome are 5, 15, 25, ... there are 10 favourable outcome

Therefore, P(該數能被５整除，已知該數是奇數) = 10/50 = 1/5


(b)

Similarly,

~ 已知該數是4的倍數: so possible outcome are 4, 8, 12, 16... there is only 25 possible outcome (so 分母 is 25, NOT 100).

~ 該數能被10整除: WITHIN those 25 choices, possible outcome are 20, 40, 60, ... there are 5 favourable outcome

Therefore, P(該數能被10整除，已知該數是4的倍數) = 5/25 = 1/5

2007-12-10 10:23:06 補充：
wrong answer... sorry

1) P(一個男孩和一個女孩)
= 黃太三個兒子 X 李太三個女兒 + 黃太二個女兒 X 李太二個兒子
= 3/5 X 3/5 + 2/5 X 2/5
= 13/25

2) (a)
P ( 該數能被５整除，已知該數是奇數 )
= 該數能被５整除 X 該數是奇數
= 20/100 X 50/100
= 1/10

2) (b)
P (該數能被10整除，已知該數是4的倍數)
= 該數能被10整除 X 該數是4的倍數
= 10/100 X 25/100
= 1/10 X 1/4
= 1/40

2007-11-24 16:15:34 補充：
第一條一定唔係上面果位人兄既答案因為要計黃太既兒子 李太既女兒 = 一個男孩和一個女孩AND黃太既女兒 李太既兒子 = 一個男孩和一個女孩

2007-11-24 16:16:24 補充：
無晒 d /加/ 號 ..第一條一定唔係上面果位人兄既答案因為要計黃太既兒子 加 李太既女兒 = 一個男孩和一個女孩AND黃太既女兒 加 李太既兒子 = 一個男孩和一個女孩