probability stats

8/9/2019 Probability stats

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PROBABILIT

Y


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THEOREMS of PROBABILITY:

Theorem 1. Probability is expressed as a numberbetween 1.000 and 0, where a value of 1.000 is acertainty that an event will occur and a value of 0 is acertainty that an event will not occur.

Theorem 2. If P(! is the probability that event willoccur, then the probability will not occur, P(!, is 1" P(!.

#ample Problem$ If the probability of %ndin& an error on an incometax return is 0.0' what is the probability of %ndin& an acceptable

return

#olution$ P(! ) 1.000 " P(! ) 1.000 * 0.0'0 )0.+0


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When to Use Theorems 3, 4, 6 an !

-ne event Two or ore /vents

-ut of Two or -ut of Two or ore

ore /vents /vents

********************************* ******************************

Theorem Theorem ' Theorem Theorem 3

utually4ot utually Independent 5ependent/xclusive /xclusive


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Theorem . If and 6 are mutually exclusive events,then the probability that either event or event 6will occur is the sum of their respective probabilities.P( or 6! ) P(! 7 P(6!

utually exclusive means that the occurrence of one eventma8es the other event impossible. Thus, if one throw of a die a occurred (event !, then event 6, say a 9, could not possiblyoccur.

4ote$ :henever an ;or< is verbali=ed, the mathematicaloperation is a"t"on, or as we shall see in Theorem ', it canbe subtraction. Theorem was illustrated with two events " it ise>ually applicable for more than two ?P( or 6 or @ or A! ) P(!7 P(6! 7 @ 7 P(A!B. Theorem is fre>uently referred to as theadditive law of probability.


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Sam#$e Pro%$em: Ta%$e %e$o& sho&s Ins#e't"onRes($ts %) S(##$"er

#upplier 4o. cceptable 4o. 4onconformin& Total

C 90 9

D 129 11

E 392 33

Total 290 11 21

If the 21 parts described in the Table are contained in a box, what is theprobability of selectin& a random part produced by supplier C or bysupplier E

#olution$ P(C or E! ) P(C! 7 P(E! ) (9 7 33!F21) 0.'+G

:hat is the probability of selectin& a nonconformin& part from supplier Cor an acceptable part from supplier E

P(nc C or ac E! ) P(nc C! 7 P(ac E! ) ( 7 39!F21 ) 0.2++


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Theorem '. If event and event 6 are not mutually

exclusive events, then the probability of either event or event 6 or both is &iven by P( or 6 or both! )P(! 7 P(6! * P(both!

4ote$ /vents that are not mutually exclusive havesome outcomes in common.

#ample Problem$ If the 21 parts described above are contained

in a box, what is the probability that a randomly selected part

will be from supplier C or a nonconformin& unit #olution $ P(C or nc or both! ) P(C! 7 P(nc! " P(C and

nc! ) (9 7 11 " !F21 ) 0.2'


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Henn 5ia&ram for Theorem ' of the above sample


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Theorem 9. The sum of the probabilities of the events of asituation is e>ual to 1.000. P(! 7 P(6! 7 @ 7 P(4! )1.000

#ample Problem$ health inspector examines three products in asub&roup to determine if they are acceptable. Arom past experienceit is 8nown that the probability of %ndin& nonconformin& units in the

sample of is 0.++0, the probability of 1 nonconformin& unit in thesample of is 0.00, and the probability of %ndin& 2 nonconformin&units in the sample of is 0.00. :hat is the probability of %ndin& nonconformin& units in the sample of

#olution$ There are only ' events to this situation$ 0 nc

unit, 1 nc unit, 2 nc units nc units.P(0! 7 P(1! 7 P(2! 7 P(! ) 1.000

0.++0 7 0.00 7 0.00 7 P(! ) 1.000

P(! ) 0.001


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Theorem . If and 6 are independent events, then theprobability of both and 6 occurrin& in the product of theirrespective probabilities. P( and 6! ) P(! 7 P(6!

4ote$ n independent event is one where its occurrence has no inJuenceon the probability of the other events or events. This theorem is referredto as the multiplicative law of probabilities. :henever an ;and< isverbali=ed, the mathematical operation is multiplication.

#ample Problem$ If the 21 parts described in the above Table arecontained in a box, what is the probability that two randomly selected

parts will be from supplier C and supplier D ssume that the %rst part isreturned to the box before the second part is selected (called withreplacement!.

#olution$ P(C and D! ) P(C! K P(D! ) (9F21!(11F21! ) 0.102


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Theorem 3. If and 6 are dependent events, theprobability of both and 6 occurrin& is the product of theprobability of and the probability that if occurred, then6 will occur also. P( and 6! ) P(! x P(6!

4ote$ The symbol P(6! is de%ned as the probability of event 6provided that event has occurred. dependent event is one whoseoccurrence inJuences the probability of the other event or events. Thistheorem is referred to as the conditional theorem, since the probabilityof the second event depends on the result of the %rst event. It isapplicable to L2 events.

#ample Problem$ ssume that in the precedin& problem the %rst partwas not returned to the box before the second part is selected. :hatis the probability

#olution$ P(C and D! ) P(C! x P(DC!) (9F21!(11F20! ) 0.102


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:hat is the probability of both parts from supplier E

#olution$ P(E and E! ) P(E! x P(EE!

) (33F21!(3F20! ) 0.0G

#ample Problem usin& theorem and $ :hat is theprobability that two randomly selected parts (withreplacement! will have one acceptable fromsupplier C and one acceptable from supplier D or

supplier E

#olution$ P?ac C and (ac D or ac E!B) P(ac C!?P(acD! 7 P(ac E! ) (90F21!(129F21739F21! )0.1'3


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/xercises$1. If an event is certain to occur, what is its probability If an

event will not occur, what is its probability2. :hat is the probability that you will live forever :hat is the

probability that an octopus will Jy

. If the probability of obtainin& a on a *sided die is 0.13,what is the probability of obtainin& any number but a

'. 5etermine an event that has a probability of 1.000.9. The probability of drawin& a pin8 chip from a bowl of

diMerent*colored chips is 0.9, the probability of a blue chipis 0.', the probability of a &reen chip is 0.19, and theprobability of a purple chip is 0.0'. :hat is the probability of

a blue or a purple chip :hat is the probability of a pin8 or ablue chip

. t any hour in a hospital intensive care unit the probabilityof an emer&ency is 0.2'3. :hat is the probability that there

will be tran>uility for the staM


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3. If a hotel has 20 8in&*si=e beds, 90 >ueen*si=e beds, 100

double beds, and 0 twin beds available, what is theprobability that you will be &iven a >ueen*si=e or a twin bedwhen you re&ister

G. ball is drawn at random from a container that holds Gyellow balls numbered 1 to G, oran&e balls numbered 1 to

, and 10 &ray balls numbered 1 to 10. :hat is theprobability of obtainin& an oran&e ball or a ball numbered 9or an oran&e ball numbered 9 in a draw of one ball :hat isthe probability of a &ray ball or a ball numbered G or a &rayball numbered G in a draw of one ball

+. If the probability of obtainin& 1 nonconformin& unit in asample of 2 from a lar&e lot of neoprene &as8ets is 0.1G andthe probability of 2 nonconformin& units is 0.29, what is theprobability of 0 nonconformin& units

10. Nsin& O+ problem, %nd the probability of obtainin& 2

nonconformin& units on the 1st

sample of 2 and 1nonconformin& unit on the 2ndsample of 2. :hat is the


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AN45/4T PQI4RIP/ -A

R-N4TI4SIf an operation can be performed in n1

ways and it has been done, a 2ndoperation can be performed in n

2

ways

and it has been done , the rd operationcan be performed in nways and so on ,

then all these operations can be

performed in n1 n2 n ....n8


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/CP/1. In how many ways can one order a bottle ofsoftdrin8s and a sandwich if a canteen oMers G

diMerent softdrin8s and 3 diMerentsandwiches

#olution $ G x 3 ) 9 ways

2. Uow many 2 di&its numbers can be formedfrom 1 , 2 , , 9 if the number formed is odd

#olution $ since odd number $ 1 , , 9 ) waysthen the tens place can be any of the ' di&its

therefore $ ' x ) 12 diMerent ways


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/CP/'. In how many ways can coins fall

#olution $ a coin can fall in 2 ways $ U , T

coins can fall in $ 2 x 2 x 2 ) G

9. Uow many possible outcomes will therebe in tossin& 2 dice

#olution $ die can fall in ways

2 dice can fall in x ) ways


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P/QNTTI-4" an arran&ement of n diMerent obVects.

a permutation of a no. of obVects in an

arran&ement of all parts of these obVects in ade%nite order.


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The permutation of n distinct obVects ta8en r at a time is&iven by$

nPr ) n(n*1!(n*2!......(n*r71!

) W if r ) n ) nX

or$ if r ) n, the number of permutations of n distinct obVects is$

nPn ) n(n!(n*1!(n*2!......1 ) nX

nX " read as n factorial

X ) x 2 x 1 )

2X ) 2 x 1 ) 2

'X ) ' x x 2 x 1 ) 2'

0X ) 1


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/CP/1. In how many ways can 3 people be

seated in a row if only ' chairs areavailable

#olution$The total number of possiblearran&ement of 3 people ta8en ' at a

time is nPr ) ) )

)


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/CP/2. In how many ways can ' boys and &irls

seated in a row of 9 chairs

. n encyclopedia consists of 12 volumes,no. 1 to 12. In how many ways can thesebe arran&ed on a shelf not necessarily inorder

#olution$ nPn ) 12P12) 12X ) nX

12X ) 12 x 11 x 10 x + x G x 3 x x9 x ' x x 2 x 1

) '3+ 001 00

#olution$P9) ) ) 3 x x 9 x ' x )

2920


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4-T/If amon& n obVects, n1are identical, n2others are identical, still n others areidentical, the no. of distinct permutationsof the obVects ta8en alto&ether is$


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/CP/

1. Uow many distinct permutations can beformed from the letters of the word#TTI#TIR#

#olution$

etters #, T appears times, I appearstwice

P ) )

) ) 90,'00


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/CP/2. In how many ways can 1 balloons be

placed in one after the another if ' are

yellow, 2 are blue, are red and the restare &reen

#olution$

P ) ) 29,229,200


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RIQRNQ P/QNTTI-4

the number of permutations of n distinctobVects arran&ed in a circle is$

(n " 1!X

/xample$In how many ways can 10 people sit in around table if they can sit anywhere

#olution$

(n*1!X ) (10 " 1!X ) +X

+X ) + x G x 3 x x 9 x ' x x 2 x1 ) 2,GG0


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R-6I4TI-4##uppose we have ' di&it numbers$ 1, 2, , '

we 8now that there are 2' diMerent permutationsof ' diMerent numbers ta8en at a time.

without re&ard to the order, there are only ' waysin which the can be chosen from the '. These

are called 'om%"nat"ons.

The number of n obVects ta8en r at a time isdenoted by nRr. In the &iven example we %nd that

for each combination, there are X -r diMerentpermutations, therefore the total number ofpermutations can be written as$

'P) 'R X


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In &eneral$

nPr ) nRr rX

nPr ) nRr rX (divide by rX!

W nPr )

nRr )


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/CP/1. In how many ways can a committee of '

be chosen from a &roup of G people

#olution$

nRr )

GR' ) ) ) 30


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/CP/2. In how many ways in which we can select 2

spades and diamonds from a dec8 of cards.

#olution $ The number of ways in which wecan select 2 from 1.

1R2) )

diamonds from 1

1R) ) 2G6y the fundamental principle of countin&

1R2 1R) (3G! (2G! ) 22,0G


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/CP/. box contains 9 red , ' blue and white

balls. In how many ways can we select balls such that$

a. ll are diMerent colorsb. ll are red

c. Two blue , one white

d. /xactly 2 are blue

e. 4one is blue


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a. 9R1 'R1 R1

)

) 9 x ' x ) 0

b. 9R) ) 10

c. 'R2 R1) x ) 1G

d. 'R2 GR1) x G ) 'G

e. GR)


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5I#RQ/T/ PQ-66IITD6I4-I$ * population is in%nite

* deals with 2 events (successFfailures, &oodFbad, etc!

* probability is constant

* probability is independent

P*+ - n.+*#+*/0#n0+

P-I##-4$ * n is lar&e (&reater than 0! probability is small * probability is constant

* probability is independent

P*+ - *1+e012+ - *n#+e0n#52+

UDP/QS/-/TQIR$ * population is %nite rather small * probability is not constant

* samplin& is without replacement

P*+ - *r.+ 70r.n0+27.n


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#P/# #-NTI-4#1! sample of 9 is selected from a steady stream of product

and proportion of 4R is 10Y.a! :hat is the probability of 14R in the sample

b! :hat is the probability of 1 or less

c! :hat is the probability of 2 or more

#olution$ a! P(1! ) 9R1(.1!1(1*.1!9*1 ) 893;

b! P(1 or less! ) P(1! 7 P(0!

) 0.2G 7 9R0(.1!0(1*.1!9*0 ) 89


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2! If 2Y of the bulbs produced by a company are defective, determine

that in the sample of 0 bulbs (a! bulbs are defective

b! not more than bulbs are defective

c! at least 2 bulbs are defective

#olution$ a! P(! ) ?(0K0.02!e*1.2BFX ) 898;6!

b! P(not L! ) P(! 7 @7 P(0!

) 0.0G3 7 ?(0K0.02!2e*1.2BF2X 7 ?(0K0.02!1e*1.2BF1X

7 ?(0K0.02!0e*1.2BF0X ) 89


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! batch of + resistors has reVect units, what is

the probability of $ a! 1 reVect in a sample of '

b! 2 reVect in a sample of '

c! ' reVect in a sample of '

#olution$ a! where$ 4) n)' r)

P(1! ) (R1 K+*R'*1!F+R' )0.'3

b! P(2! ) (R2 K+*R'*2!F+R' )0.93

c! Impossible, or it will not happen, since there are

only reVects in the population.


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7ORMAL PROBABILITY =ISTRIBUTIO7

or R-4TI4N-N# PQ-66IITD

* 8nown also as Saussian 5istribution * the most si&ni%cant probability distribution in

the entire theory of statistics

* &raphically represented by symmetrical bellshaped curve 8nown as the normal curve


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RUQRT/QI#TIR#

* mean, median, mode " same value

* curve is symmetrical about the vertical

line which contains the mean * curve is asymptotic to the hori=ontal axis

(tail is rarely extended beyond ' or 9 sdfrom the mean

* total area under the curve is e>ual to one

(1.000!


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R-4TI4N-N# PQ-66IITDIn practice, we %nd areas under the &raph of a normal distribution or

normal curve by performin& a simple chan&e of scale, in which we

convert the units of measurement in the ori&inal, or x*scale into standard

units, or =*scores by means of this formula$ > - *+012?

In this new scale, = simply tells us how many standard deviations the

correspondin& x*value lies above or below the mean of its distribution

/xample 1$ Aind the area under the curve between 12 19 with

a! Z ) 10 and [ ) 9 b! Z ) 20 and [ ) 10

#olution$ a! = )(12*10!F9)0.'0 and =) (19*10!F9)1.00

thus, = ) 0.'1 * 0.199' ) 0.1G9+

b! = )(12*20!F10)*0.G0 and =) (19*20!F10)*0.90

thus, = ) 0.2GG1 * 0.1+19 ) 0.0+

(Q/A/Q T- T6/ -A 4-Q RNQH/ A-Q TU/ HN/#$0.'1, 0.199', 0.2GG1, /TR!


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/xample 2$ Aind the area under the standard

normal curve between = ) *1.20 = ) 0#olution$ from the table *1.20 ) 0.G'+

/xample $ If a random variable has the standard

normal distribution, what are the probabilities that

it will ta8e on a value$

a! less than 1.' d! less than *1.9

b! &reater than *0.'3 e! between 0.+9 1.

c! &reater than 0.3 f! between *0.'9 0.9

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