Simple Probability

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.

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)

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