28-17

nicboy

    

Q17:

    

If a certain coin
is flipped, the probability that the coin will land heads is 1/2.  If the coin is flipped 5 times, what is the
probability that it will land heads up on the first 3 flips and not on the last
2 flips?

    

 

    

A.  3/5

    

B.  1/2

    

C.  1/5

    

D.  1/8

    

E.  1/32

    

Answer:

请教思路是怎样的。谢谢

bth

"If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips" ...

Method-1:
When a coin is flipped 5 times, there are 2^5 = 32 possible combinations of outomes ... out of those 5c3 = 10 combos would have 3 Heads & 2 Tails.

i.e., 5c3/(2^32) = 10/32 = 5/16

Method-2:
Probability of landing 3 Heads out of 5 Tosses = 5c3 * (probability of head)^3 * (probability of tail)^2
i.e., 5c3 * (1/2)^3 * (1/2)^2
i.e., 5c3 * (1/2)^5 = 10/32 = 5/16

What's the OA?

skidrow

e 吧 （1/2）^5

bth

Yes,

Flip1:  Head: 1/2

Flip2: Head: 1/2

Flip3: Head: 1/2

Flip4: Tail: 1/2 So, Head: 1/2

Flip5: Tail: 1/2 So, Head: 1/2

Total: 1/2*1/2*1/2*1/2*1/2=(1/2)^5 =1/32.

IMO: E.

nicboy

I don't understand what is 5C3 or you mean C(3,5)?
What's the answer, 5/16 or 1/32

[此贴子已经被作者于2008-6-22 21:13:54编辑过]

bth

Answer is 1/32, no doubt.

kevintown

the official answer on TT is D! I don't understand.

it should be 5/16, I've done several this kind of question in Feifei

纽约客

应该是5/16.  这道题是一道标准的柏努利分布。计算如下：(5!/(3!*2!))*0.5^3*(1-0.5)^2

有关柏努利的知识，请见下面：

    

Bernoulli Processes and the Binomial Distribution

    

Example: An airline reservations
switchboard receives calls for reservations, and it is found that when a
reservation is made, there is a good chance that the caller will actually show
up for the flight. In other words, there is some probability p (say for now p = 0.9) that the caller will show up and buy the ticket the day of
departure.

    

Consider
a single person making a reservation. This particular reservation can either
result in the person on the flight (a success)
or a “no show” (a failure). Let X (a random variable) represent the
result of a particular reservation. That is, we could assign a value of 1 to X
if the person shows up for the flight (X
= 1), and let X = 0 if the person
does not. Then,

    

P(X
= 0) = 1 - p   and   P(X
= 1) = p.

    

This
is an example of a Bernoulli process,
named for the Swiss mathematician James Bernoulli (1654-1705). A Bernoulli
process is a sequence of n identical
trials of a random experiment such that each trial:

    

(1)   produces one of two possible
complimentary outcomes that are conventionally called success and failure and

    

(2)   is independent of any other trial so
that the probability of success or failure is constant from trial to trial.

    

Note
that the success and failure probabilities are assumed to be constant from
trial to trial, but they are not necessarily equal to each other. In our
example, the probability of a success is 0.9 and the probability of a failure
is 0.1.

    

    
    

The
number of successes in a Bernoulli process is a binomial random variable.
The probability that a binomial variable X
will take on any particular value x is
given by the binomial formula:

    

        
            
                    
                    
                
                
                    
                    
                
            
    

    

If
X is a Binomial(n, p) random variable,
then

    

        
            
                    
                    
                    
                
                
                    
                    
                    
                
                
                    
                    
                    
                
            

Expected   Value:	E(X)	=   np
Variance:	Var(X)	=   np(1 - p)
Standard   Deviation:	s(X)	=

    

    

jh1904

题目是前三次head, 后二次tail，而不是三次head, 二次tail的组合。答案应该是1/32。

如果是问“三次head, 二次tail”的机率，那才是Bernoulli Process