1. Each of the 3 letters is to be addressed to David , Joan and Peter respectively.
The letters are randomly put into 3 envelops each with one of their names.
Find the probability that
(i)only John wii get the right letter
(ii)none of the letters goes into the right envelopes
(iii)all letters go into the right envelopes
2. Both Mr.Wong and Mrs Chan have 2 children.
Find the probability that he has 2 boys and she has 2 girls.
Simple Probability
2011-01-27 6:10 am
回答 (3)
2011-01-27 7:50 pm
✔ 最佳答案
I suggest you to use Tabulation or a Tree Diagram to solve it.Or you can simply list out all possibilities in this question for the number of outcomes is small.
Denote David, Joan and Peter by D, J and P resp.
There are 6 possibilities:
( D, J, P ): All correct
( D, P, J ): Only D is correct
( J, D, P ): Only P is correct
( J, P, D ): All wrong
( P, D, J ): All wrong
( P, J, D ): Only J is correct
1 (i) P( John correct ) = 1/6
====================
1 (ii) P( No correct ) = 2/6 = 1/3
=======================
1 (iii) P( All correct ) = 1/6
================
2. P( Mr. Wong has 2 sons, Mrs. Chan has 2 daughters )
= ( 1/2 )( 1/2 )( 1/2 )( 1/2 )
= 1/16
=====
Hope I can help you.
2011-01-31 22:18:58 補充：
Thanks 魔龍帝齊格益力多.
I guess he is a Form 3 student, so I use listing instead of those S.5 Multiplication Laws to answer his question.
I am sure he can understand in this way.
參考: Mathematics Teacher Mr. Ip
2011-01-28 2:49 am
Although the method that Superman used is a slow way,but the format of 002
is wrong.
It should be:
P(only John wii get the right letter)
is wrong.
It should be:
P(only John wii get the right letter)
2011-01-27 10:48 pm
其實第一題唔使咁煩、咁長篇大論：
1. (i)(1/3)X(1/2)X(1/1)=1/6
(ii) (2/3)X(1/2)X(1/1)=1/3
(iii) (1/3)X(1/2)X(1/1)=1/6
至於第二題：
2. [(1/2)X(1/2)]X[(1/2)X(1/2)]=1/16
1. (i)(1/3)X(1/2)X(1/1)=1/6
(ii) (2/3)X(1/2)X(1/1)=1/3
(iii) (1/3)X(1/2)X(1/1)=1/6
至於第二題：
2. [(1/2)X(1/2)]X[(1/2)X(1/2)]=1/16
收錄日期: 2021-04-13 17:47:58
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