ch2

ch2ch2their distributionstheir distributions

Random variables andRandom variables and

Random variablesRandom variables

DistributionsDistributions

Some Important Discrete 2.3

Discrete Probability Distributions

Probability Distributions

nxxx ,,, 21

nin

xpxXP ii ,,2,1,1

)()(

)1

;(~n

xfX i

Uniform

We have a finite set of outcomes

which has the same probability of occurring (equally

likely outcomes).

X is said to have a Uniform distribution and we

each of

Write

So

)1

;(n

xf ibulb,

Example 2.2When a light bulb is selected at random from a boxthat contains a 40-watt bulb, a 60-watt bulb, a 75-watt

and a 100-watt bulb. Find

Solution:

.6,5,4,3,2,1,6

1)6;( xxf

Example 2.3

When a die is tossed, S={1,2,3,4,5,6}.

P(each element of the sample space) = 1/6.

Therefore, we have a uniform distribution, with

Bernoulli trial

A Bernoulli trial is an experiment which has two

Let p = P(success), q = P (failure ) (q=1-p).

‘success’ and ‘failure’.possible outcomes:

0)(1)( failureXandsuccessX

0,1

1,)()(

xpq

xpxpxXPThe pmf of X is

Xkp

0p1

1p

or

伯努利资料

Binomial

each of which must result in either a ‘success’ with

Consider a sequence of n independent Bernoulli trials

probability of p or a ‘failure’ with probability q=1-p.

Let X= the total number of successes in these n trial}.,1,0{ nRX so that

nxppCxXP xnxxn ,,1,0,)1()(

X is said to have a Binomial distribution with parameters

P( the total number of x successes ) =

n and p and we write X~Bin(n, p) or X~b(x;n, p)

1,0)1()( xppxXP xnx

),1( pb

X is said to have a Binomial distribution with parameters

n and p and we write X~Bin(n, p) or X~b(x;n, p)

Special case,

when n=1,we have

We write

B(n,p)1n

b(1, p)

二项分布的图形

The probability that a certain kind of component will survive a given shock test is 3/4. Find the probability that exactly 2 of the next 4 components tested survive.

Example 2.4

Example 2.5

The probability that a patient recovers from a rare blood

disease is 0.4. this disease,

survive,

If 15 people are known to have contracted

what is the probability that (a) at least 10 (b) from 3 to 8 survive, and (c) exactly 5 survive?

Solution:

Solution:

from a shipment.

(b) Suppose that the retailer receives 10 shipments in a

A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.

Example 2.6

(a) The inspector of the retailer randomly picks 20 items

will be at least one defective item among these 20?What is the probability that there

shipments containing at least one defective device?shipment. month and the inspector randomly tests 20 devices per

What is the probability that there will be 3

Solution ：

,2,1,0,!

)( xx

exXP

x

Poisson

The pmf of a random variable X which has a Poisson

0distribution with parameter is given by

).;(~ xpXand we write 泊松资料

电话呼唤次数 交通事故次数商场接待的顾客数

地震 火山爆发 特大洪水

泊松分布的图形

单击图形播放 / 暂停　 ESC 键退出

二项分布 泊松分布)( nnp

During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is the probability that 6 articles enter the counter in a given millisecond?

Example 2.7

Solution ：

handle at most 15 tankers per day.

Example 2.8

Ten is the average number of oil tankers arriving each day a certain port city. The facilities at the port can

What is the probability that on a given day tankershave to be turned away? Solution ：

Solution: }100,75,60,40{S

each element of the sample space occurs with

probability 1/4.

Therefore, we have a uniform distribution:

,4

1)4;( xf .100,75,60,40x

4

3,4;2b 222

4 )4

31()

4

3( C 4

2

4

3

!2!2

!4

Assuming that the tests are independent andSolution:

p=3/4 for each of the 4 tests, we obtain

128

27

)10( XP

9662.01

)83( XP

9

0

)4.0,15;(1x

xb)10(1 XP

0338.0

0.0271-0.9050

8

3

)4.0,15;(x

xb

8

0

2

0

)4.0,15;()4.0,15;(x x

xbxb

Solution ：Let X = the number of people that survive.

(a)

(b)

8779.0

Solution ：

(a) Denote by X the number of defective

Devices among the 20.Then this X follows a b (x; 20,0.03). Hence

0200 )03.01(03.01 )1( XP

4562.0

)03.0,20;0(1 b)0(1 XP

(b) Assuming the independence from shipment to shipment and denoting by Y .

Y=the number of shipments containing at least one

defective.Then Y~b(y;10,0.4562).

.1602.03103310 )4562.01(4562.0 C)3( YP

Therefore,

)4;6(p

4

7851.08893.0

!6

464

e

Solution ：Using the Poisson distribution with x=6 and

, We find from Table 1 that

.1042.0

6

0

5

0

)4;()4;(x x

xpxp

Solution ：Let X be the number of tankers arriving eachday.

Then,

)15( XP

0487.09513.01

15

0

)10;(1x

xp)15(1 XP

using Table, we have

Jacob Bernoulli

Born: 27 Dec 1654 in Basel, Switzerland

Died: 16 Aug 1705 in Basel, Switzerland

伯努利资料

泊松资料

Born: 21 June 1781 in Pithiviers, France

Died: 25 April 1840 in Sceaux (near Paris), Fr

ance

Siméon Poisson