session 4

2
7/3/2012 1 How reliable is Reliable? Let D denote the event that an IC is defective Let D R denote the event that ‘reliable’ test finds the IC to be defective Given, P(D) = .005 P[D R | D] = 198/200 = 0.99 P [D R | not D] = 8/160 = 0.05 P[D R ] = P[D R and D] + P[D R and not D] = P[D] × P[D R | D] + P[not D] × P[D R | not D] = 0.005 × 0.99 + 0.995 × 0.05 = 0.00495 +0.04975 = 0.0547 P[D| D R ] = P[D R and D] / P[D R ] = 0.00495/0.0547 = 0.0905 P[not D| D R ] = 0.9095 Same calculation using table Events Prior Prob (i) P[DR|..] (ii) P[DR and ..] (iii) =(i)*(ii) P[..|DR] (iv) =(iii)/P[DR] D 0.005 0.99 0.00495 P[D| DR] =0.0905 Not D 0.995 0.05 0.04975 P[not D| DR] =0.9095 Sum 1 P[DR]= 0.0547 If Reliable finds the IC to be not defective Events Prior Prob (i) P[not DR|..] (ii) P[not DR and ..] (iii) P[..|not DR] D 0.005 0.01 0.00005 P[D| not DR] =0.000053 Not D 0.995 0.95 0.94525 P[not D| not DR] =0.999947 Sum 1 P[not DR]= 0.9453 Bayes Rule ] [ ] [ ] | [ ] [ ] | [ ] [ ] [ c B P B A P A B P B A P A B P A P B A P + × = = ] | [ ] [ ] | [ ] [ ] | [ ] [ c c A B P A P A B P A P A B P A P × + × × = Bayes Rule -- using table events Prior Prob P[A i ] (i) P[B|Ai] (ii) P[BAi] (iii) =(i)*(ii) P[Ai|B] (iv) =(iii)/P[B] A 1 P[A 1 ] P[B|A1] P[A1B] P[A 1 | B] A 2 P[A 2 ] P[B|A2] P[A2B] P[A 2 | B] A k P[A k ] P[B|Ak] P[AkB] P[A k | B] Sum 1 P[B]

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Page 1: Session 4

7/3/2012

1

How reliable is Reliable?

Let D denote the event that an IC is defective

Let DR denote the event that ‘reliable’ test finds the IC to be defective

Given, P(D) = .005 P[DR| D] = 198/200 = 0.99

P [DR| not D] = 8/160 = 0.05

P[DR] = P[DR and D] + P[DR and not D]

= P[D] × P[DR| D] + P[not D] × P[DR| not D]

= 0.005 × 0.99 + 0.995 × 0.05 = 0.00495 +0.04975 = 0.0547

P[D| DR] = P[DR and D] / P[DR] = 0.00495/0.0547 = 0.0905

P[not D| DR] = 0.9095

Same calculation using table

Events Prior Prob

(i)

P[DR|..]

(ii)

P[DR and ..]

(iii)

=(i)*(ii)

P[..|DR]

(iv)

=(iii)/P[DR]

D 0.005 0.99 0.00495 P[D| DR]=0.0905

Not D 0.995 0.05 0.04975 P[not D| DR]=0.9095

Sum 1 P[DR]=

0.0547

If Reliable finds the IC to be

not defective

Events Prior Prob

(i)

P[not DR|..]

(ii)

P[not DRand ..]

(iii)

P[..|not DR]

D 0.005 0.01 0.00005 P[D| not DR]=0.000053

Not D 0.995 0.95 0.94525 P[not D|

not DR]=0.999947

Sum 1 P[not DR]=

0.9453

Bayes Rule

][][

]|[][]|[

][

][

cBP

BAP

ABPBAP

ABPAPBAP

∩+∩

×==

]|[][]|[][

]|[][cc ABPAPABPAP

ABPAP

×+×

×=

Bayes Rule -- using table

events Prior Prob

P[Ai]

(i)

P[B|Ai]

(ii)

P[BAi]

(iii)

=(i)*(ii)

P[Ai|B]

(iv)

=(iii)/P[B]

A1 P[A1] P[B|A1] P[A1B] P[A1| B]

A2 P[A2] P[B|A2] P[A2B] P[A2| B]

Ak P[Ak] P[B|Ak] P[AkB] P[Ak| B]

Sum 1 P[B]

Page 2: Session 4

7/3/2012

2

First Can contains 10 marbles: 7 red & 3 blue

Second Can contains 4 blue & 1 red

1 marble 1 marble

1 marble

P(the marble drawn from the

Box is blue)=?

The marble drawn from

the Box is found to be

blue. What is the prob.

that it came from Can2?

Monty Hall Problem

or

‘Khul Ja SimSim’

• The car is behind one of the 3 doors.

• You select door A.

• Aman Verma (who knows where the car is)

opens door B and shows that this is empty.

Gives you an option of “switch”.

• Should you stick to your initial choice?

Solution to KJSS

Correct

Door

Prior

Probabi

lity

P(door B is opened

given you select

door A and correct

door is..)

P( correct door is /

and door B is

opened and you

selected door A)

P( correct door is /

given Door B is

opened and you

selected door A)

A 1/3 1/2 1/6 1/3

B 1/3 0 0 0

C 1/3 1 1/3 2/3

P(door B is

opened) 1/2

Will the new product do well?Combine prior opinion with pilot survey

Will do well = 40% of potential customers will like it

Won’t do well = 20% will like it

PRIOR PROB

60%

40%

Pilot survey: from 25 potential customers

X like it

Want P[will do well | X out of 25 like it]

AB

Is it easier to find

P[B|A] ?