Probability

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					Law of Total Probability
   and Bayes’ Rule
   “Event-composition method”
• Understand the experiment and sample
  points.
• Using set notation, express the event of
  interest in terms of events for which the
  probability is known.
• Applying probability rules, combine the
  known probabilities to determine the
  probability of the specified event.
             Problem 2.86
• In a factory, 40% of items produced come
  from Line 1 and others from Line 2.
• Line 1 has a defect rate of 8%.
  Line 2 has a defect rate of 10%.
• For randomly selected item, find probability
  the item is not defective.
     A: the selected item is not defective
                Problem 2.86
• A: the selected item is not defective.

                          S
            A


       B1            B2       A  ( A  B1 )  ( A  B2 )

• B1: item came from Line 1.
  B2: item came from Line 2.
               Problem 2.86
• So we may write A  ( A  B1 )  ( A  B2 )
• Since this is the union of disjoint sets,
  the Additive Law yields
       P( A)  P( A  B1 )  P( A  B2 )
• Or, in terms of conditional probabilities
    P( A)  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )

   P( A)  (0.08)(0.40)  (0.10)(0.60)  0.092
         The Decision Tree
                 defective

Line 1
                not defective



Line 2           defective



                not defective
               Problem 2.94
• Must find blood donor for an accident victim in
  the next 8 minutes or else…
• Checking blood types of potential donors requires
  2 minutes each and may only be tested one at a
  time.
• 40% of the potential donors have the required
  blood type.
• What is the probability a satisfactory blood donor
  is identified in time to save the victim?
             Finding a Donor
• A: blood donor is found within 8 minutes
• Some sample points: “B bad, G good”
  A = { (G), (B,G), (B,B,G), (B,B,B,G) }
• Let Ai: ith donor has correct blood type
   A  ( A1 )  ( A1  A2 )  ( A1  A2  A3 )
                         ( A1  A2  A3  A4 )

           4 mutually exclusive events
                   Finding a Donor
A  ( A1 )  ( A1  A2 )  ( A1  A2  A3 )  ( A1  A2  A3  A4 )

 • Trials are independent and each P(Ai) = 0.40,
   and so
      P( A)  P( A1 )  P( A1 ) P( A2 )  P( A1 ) P( A2 ) P( A3 )
                 P( A1 ) P( A2 ) P( A3 ) P( A4 )

      P( A)  0.4  (0.6)(0.4)  (0.6)(0.6)(0.4)
                  (0.6)(0.6)(0.6)(0.4)
              .8704
                     Finding a Donor
            A1         saved!

                            A2         saved!

            A1                               A3   saved!

                           A2                        A4    saved!
                                             A3
      Or, more simply,
P( A)  1  P(donor is not found)                    A4     too
      1  P( A1 ) P( A2 ) P( A3 ) P( A4 )                  late!

      1  (0.6) 4
              Problem 2.96
• Of 6 refrigerators, 2 don’t work.
• The refrigerators are tested one at a time.
• When tested, it’s clear whether it works!
A. What is the probability the last defective
   refrigerator is found on the 4th test?
B. What is the probability no more than 4 need
   to be tested to identify both defective
   refrigerators?
              Problem 2.96

C. Given that exactly one defective
   refrigerator was found during the first 2
   tests, what is the probability the other one
   is found on the 3rd or 4th test?
Partition of the Sample Space
                                        S


        B1          B2     …       Bk

 A collection of sets {B1 , B2 ,    , Bk } such that
 1. S  B1  B2          Bk , and
 2. Bi  B j  , for i  j ,
 is called a partition of S .
         Union of Disjoint Sets
                                           S
                   A

             B1         B2     …      Bk

For the partition {B1 , B2 ,   , Bk } of the sample space S ,
we may write A  ( A  B1 )  ( A  B2 )         ( A  Bk )
and so
  P(A)  P( A  B1 )  P( A  B2 )         P( A  Bk ).
           Recall Problem 2.86
• A: the selected item is not defective.

                            S
            A
       Not defective
      B1               B2       A  ( A  B1 )  ( A  B2 )

• B1: item came from Line 1.
  B2: item came from Line 2.
             Law of Total Probability
                                   S
         A
                                    If P ( Bi )  0, then
                     …
  B1          B2             Bk        P ( A  Bi )  P ( A | Bi ) P ( Bi )

Hence, for the partition {B1 , B2 ,      , Bk } of the sample space S ,
we have P(A)  P( A  B1 )  P( A  B2 )              P( A  Bk )
or equivalently,
P(A)  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )     P( A | Bk ) P( Bk ).
                  Total Probability
                              A         P(A|B1)P(B1)

             B1          A
                              A        P(A|B2)P(B2)
               B2

               B3        A
                             A          P(A|B3)P(B3)

                        A
P(A)
  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )  P( A | B3 ) P( B3 ).
     Bayes’ Theorem follows…
Since P(A)  P( A | B1 ) P( B1 )              P( A | Bk ) P( Bk ),
we also have
                     P( A  B j )
     P( B j | A) 
                        P( A)
                                   P( A  B j )
           
               P( A | B1 ) P( B1 )        P( A | Bk ) P( Bk )
 or simply,
                                   P( A  B j )
           P( B j | A)      k

                            P( A | B ) P( B )
                            i 1
                                          i        i
                               Bayes’
                                       A         P(A|B1)P(B1)

                   B1            A
                                      A         P(A|B2)P(B2)
                     B2

                     B3
                                A
                                     A           P(A|B3)P(B3)

                                A
                                     P( A  B2 )
P( B2 | A) 
             P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )  P( A | B3 ) P( B3 )
            Making Resistors
• Three machines M1, M2, and M3 produce “1000-
  ohm” resistors.
• M1 produces 80% of resistors accurate to within
  50 ohms, M2 produces 90% to within 50 ohms,
  and M3 produces 60% to within 50 ohms.
• Each hour, M1 produces 3000 resistors, M2
  produces 4000, and M3 produces 3000.
• If all of the resistors are mixed together and
  shipped in a single container,
  what is the probability a selected resistor is
  accurate to within 50 ohms?
              Making Resistors
• Define A: resistor is accurate to within 50 ohms.
• M1 produces 80% of resistors accurate to within
  50 ohms, M2 produces 90% to within 50 ohms,
  and M3 produces 60% to within 50 ohms.
        P( A | M 1 )  0.80, P( A | M 2 )  0.90,
               and P( A | M 3 )  0.60

• Each hour, M1 produces 3000 resistors, M2
  produces 4000, and M3 produces 3000.
  P( M 1 )  0.3,   P( M 2 )  0.40, and P( M 3 )  0.30.
             Using Total Probability
Since
 P(A)  P( A | M 1 ) P( M 1 )  P( A | M 2 ) P ( M 2 )  P ( A | M 3 ) P ( M 3 )
we have

         P(A)  (0.8)(0.3)  (0.9)(0.4)  (0.6)(0.3)
                  0.78

         That is,
             78 % are expected to be accurate
             to within 50 ohms.
              The Tree
          A      (0.8)(0.3)

M1                            P (A)  (0.8)(0.3)
      A
                                     (0.9)(0.4)
          A      (0.9)(0.4)
 M2                                  (0.6)(0.3)
      A                             0.78
 M3
          A      (0.6)(0.3)

      A
  …given it’s within 50 ohms…
• Determine the probability that,
  given a selected resistor is accurate to within 50
  ohms, it was produced by M1. P( M1 | A) = ?

• Determine the probability that,
  given a selected resistor is accurate to within 50
  ohms, it was produced by M3. P( M3 | A) = ?
          Given A…
          A   (0.8)(0.3)
                           P (M1 | A)
M1    A
                                  P(M1  A)
          A   (0.9)(0.4)        
 M2                                  P ( A)
                                  (0.8)(0.3)
 M3   A                         
              (0.6)(0.3)             0.78
          A

      A
                    Arthritis
• A test detects a particular type of arthritis for
  individuals over 50 years old.
• 10% of this age group suffers from this arthritis.
• For individuals in this age group known to have
  the arthritis, the test is correct 85% of the time.
• For individuals in this age group known to NOT
  have the arthritis, the test indicates arthritis
  (incorrectly!) 4% of the time.
• P( has arthritis | tests positive ) = ?
                     Arthritis
• 10% of this age group suffers from this arthritis.
               P(have arthritis) = 0.10
• For individuals in this age group known to have
  the arthritis, the test is correct 85% of the time.
     P( tests positive | have arthritis ) = 0.85

• For individuals in this age group known to NOT
  have the arthritis, the test indicates arthritis
  (incorrectly!) 4% of the time.
      P(tests positive | no arthritis ) = 0.04

• P( have arthritis | tests positive ) = ?
P(has arthritis | tests positive ) = ?

                     positive   0.85

      Has
      arthritis      negative
               0.1

               0.9
      No
      arthritis                 0.04
                     positive

                     negative
                The 3 Urns
• Three urns contain colored balls.
       Urn         Red   White   Blue
        1           3      4      1
        2           1      2      3
        3           4      3      2
• An urn is selected at random and one ball is
  randomly selected from the urn.
• Given that the ball is red, what is the
  probability it came from urn #2 ?
            The Lost Labels
• A large stockpile of cases of light bulbs,
  100 bulbs to a box, have lost their labels.
• The boxes of bulbs come in 3 levels of
  quality: high, medium, and low.
• It’s known 50% of the boxes were high
  quality, 25% medium, and 25% low.
• Two bulbs will be tested from a box to
  check if they’re defective.
               Lost Labels…
• The likelihood of finding defective bulbs is
  dependent on the bulb quality:
    Number of defects   Low   Medium   High
          0             .49     .64     .81
          1             .42     .32     .18
          2             .09     .04     .01

• Given neither bulb is found to be defective,
  what is the probability the bulbs came from
  a box of high quality bulbs?

				
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