數學概率問題... 中五

Consider the following example:

In a room, there are 3 animals: 1 white cat, 1 black cat and 1 dog.

Mr. Chan now choose one animal from the room

(Case I)
A: Choosing a cat
B: Choosing a dog

---> A and B are 互補事件

Reason: Mr. Chan cannot choose fish, monkey, snack, bear... from the room. There are only cats and dogs in the room!!! So if he doesn't choose a cat, he must have chosen a dog. That's why "Choosing a cat and choosing a dog is 互補事件", since if A doesn't happen, B must happen (vice versa)


(Case II)
R: Choosing a white cat
S: Choosing a black cat

---> R and S are 互斥事件

Reason: Mr. Chan cannot choose a white cat and a black cat AT A SAME TIME. If he has chosen a white cat, he must not have chosen a black cat at the same time. That's why "Choosing a white cat and choosing a black cat is 互斥事件", since if R happen, S must not happen (vice versa)

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IMPORTANT:
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A and B are 互補事件 means if A does not happen, B must happen (vice versa)
So P(A) + P(B) MUST EQUAL TO 1

R and S are 互斥事件 means if R happen, S must not happen (vice versa)
But P(R) + P(S) CAN BE LESS THAN 1

They are different concept!!! Be careful!!!

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Question time!!!
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Question (1): If X and Y are 互補事件, then they MUST BE 互斥事件???
Answer (1): YES!!!

Just like the example you give out, when 小明上學遲到 and 小明準時上學 are 互補事件 (their probability sum to 1), they must also be 互斥事件 (since 小明 can't 遲到 and 準時 at the same time)


Question (2): If X and Y are 互斥事件, then they MUST BE 互補事件???
Answer (2): NO!!!!!!

Just look back to case (II). P(Choosing a white cat) + P(Choosing a black cat) MAY NOT EQUAL TO 1 (since Mr. Chan can choose a dog)

So R and S is NOT 互補事件

Therefore, although R and S are 互斥事件, they are NOT 互補事件

Clear? ^_^

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Exercise:
D: Chris is a boy
E: Chris is a girl
F: Chris is 13 years old
G: Chris is 14 years old

(i) Is D and E 互斥事件?
(ii) Is D and E 互補事件?
(iii) Is F and G 互斥事件?
(iv) Is F and G 互補事件?
(v) Is D and F 互斥事件?

互斥 happens at different time,they can not happen together.
互補 means happen in same time ,their sum is equal to one.