对楼上的算法，有不同意见。答案是对的，但是方法好像有不妥。
这道题是求P（A非 n B非）。因此，（我不知道如何表达A非与B非，所以大致表示一下吧）
(1)如果A与B相互独立，则所求=P（A非）* P (B非）=（1-0.6)*(1-0.5)=0.2
（2), 如果A与B不是相互独立，如果B发生时，A总是发生。则二者都不发生的概率取决于A的概率。所求=1-0.6=0.4.
所以，A与B都不发生的概率介于0.2-0.4之间，最大概率为0.4.
如果有任何疑问，可参考如下关于计算独立概率事件与相对概率事件的讲义：
Independent Events
Two
events A and B are said to be independent if the fact that A has occurred or not does not affect your assessment of the
probability of B occurring.
Conversely, the fact that B has occurred
or not does not affect your assessment of the probability of A occurring.
1. If A
and B are independent events, then
P(A
and B) = P(A Ç
B)
= P(A) ´
P(B).
Note:
if two events E and F are independent, then E and  are also independent,
as well as  and F, and also and .
Example: What is the probability of rolling 3 sixes
in a row with one die?
Example: Oil Drilling
An
oil company decides to drill for oil in 3 separate locations. We assume:
·
The
probability of finding oil at each location is 0.10 (or a 10% chance).
·
The
event of finding oil at any location is independent
of what happens at another location.
Questions:
1) What is P(find oil in all 3 locations)?
2) What is P(find oil in exactly 2 locations)?
3) What is P(find oil in at least 1 location)?
What
are the basic outcomes of this experiment?
1)
Define some events of interest:
A1
= find oil at location 1
A2
= find oil at location 2
A3
= find oil at location 3
Then P(oil
in all 3)
|
=
P(A1
Ç
A2
Ç
A3)
|
|
=
P(A1 and A2
and A3)
|
|
=
P(A1) ´
P(A2)
´
P(A3)
|
|
=
(0.1)(0.1)(0.1)
|
|
=
0.001
|
2)
Define
B1
|
=
oil in 1 and 2 but not 3
|
B2
|
=
oil in 1 and 3 but not 2
|
B3
|
=
oil in 2 and 3 but not 1
|
Then
since B1, B2 and B3 are mutually exclusive (Are they?):
P(oil in exactly 2) = P(B1 or B2 or B3) = P(B1) + P(B2) + P(B3).
Therefore
P(B1)
= P(A1 and A2
and  ) = P(A1) ×
P(A2) ×
P().
Recall
that for an event E, P() = 1 - P(E). Therefore, P(B1) = (0.1)(0.1)(1
- 0.1) = 0.009.
Since
the probability is the same for each location P(B2) = P(B3)
= P(B1) = 0.009. Therefore,
P(oil in exactly 2) = P(B1) + P(B2)
+ P(B3) = 0.027.
3)
As for the probability of finding oil in at least one location, we could solve this
by first finding the probability of oil in exactly 1 location, then in exactly
2 locations, then in all 3, and then add these probabilities (since the events
are mutually exclusive).
But
there is an easier way by looking at the complement event. What is the
complement of this event? It is the probability of finding no oil anywhere.
P(no oil in all 3)
|
=
P( and  and )
|
|
=
P() ×
P() ×
P()
|
|
=
(0.9)(0.9)(0.9)
|
|
=
0.729.
|
So
P(oil in at least 1 location) = 1 -
0.729 = 0.271.
Conditional Probability
Example: One of the businesses that have grown out
of the public's increased use of the internet has been providing internet
service to individual customers; those who provide this service are called
Internet Service Providers (ISPs). More recently, a number of ISPs have
developed business models whereby they do not need to charge customers for
internet service at all, by collecting fees from advertisers, and forcing the
non-paying customers to view these advertisements. Jupiter Communications
estimates that eventually 20% of web users will have a free ISP. 6% of all web
users, it is estimated, will have both a free ISP and a paid ISP account.
(Assume that an internet user must have at least one ISP.)
In
2003, what proportion of internet users is expected to do the following?
a) subscribes to both a free ISP and a paid
ISP.
b) subscribes only to a paid ISP.
c) subscribes only to a free ISP.
If
a user has a paid account, what is the probability that he/she also has a free
account?
In
these simple calculations, we are making use of the conditional probability
formula:
P(A|B) = P(A holds given that B holds) = 
This
relationship is known as Bayes' Law, after the English clergyman Thomas Bayes
(1702-1761), who first derived it. Bayes' Law was later generalized by the
French mathematician Pierre-Simon LaPlace (1749-1827).
|
|
|
Bayes
|
|
LaPlace
|
Often
we make use of the equivalent formulae:
P(A
Ç
B)
|
=
P(A|B)P(B)
|
or
P(A
Ç
B)
|
=
P(B|A)P(A).
|
Example: Exactly 100 employees of a firm have each
purchased one ticket in a lottery, with the drawing to be held at the firm's
annual party. Of the 80 men who purchased tickets, 25 are single. Only 4 of the
women who purchased tickets are single.
a)
Find the probability that the lottery winner is married.
b)
Find the probability that the lottery winner is a married woman.
c)
If the winner is a man, what is the probability that he is married?
We
can use conditional probability to define independence. If two events A and B
are independent, then P(A
Ç
B) = P(A)P(B).
In terms of conditional probability, this means:
Does
this agree with our intuitive definition of independence?
Conditioning
can also help us calculate probabilities of much more complicated events.
A Useful Probability Rule: Sometimes it is easier to determine the
probability of an event A indirectly,
using another event B. This can be a
useful approach if the likelihood of A
is dependent on whether or not event B
occurs. For any events A and B,
P(A)
|
=
P(A
Ç
B) + P(A
Ç
 )
|
|
=
P(A|B)P(B)
+ P(A|)P().
|
Example: Take a deck of 52 cards. Take out 2 cards
sequentially, but don't look at the first. What is the probability the second
card you chose was a §?
| 
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发表于 2008-8-14 15:55:00
