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45-733 Probability and Statistics I (3rd Mini AY 1999-2000 Flex-Mode and Flex-Time)

Assignment#2 Answers


2.36 This is just like the men-women committee examples done in class. Clearly, there are [8 choose 4] outcomes. There are 3 undergraduates and 5 graduate students. Hence,

                         æ3öæ5ö
                         ç ÷ç ÷
                         è2øè2ø
                 P(E) = ---------
                           æ8ö
                           ç ÷
                           è4ø
2.37 Clearly, there are [52 choose 2] outcomes. There are 4 Aces and there are 12 Face cards (i.e., Jacks, Queens, Kings). Hence,

                         æ4öæ12ö
                         ç ÷ç  ÷
                         è1øè 1ø
                 P(E) = ---------
                           æ52ö
                           ç  ÷
                           è 2ø
2.38 The number of ways to divide the 9 motors between the three production lines is . There are 2 motors from a particular supplier and 7 not from that supplier. Hence, there are = ways to assign motors such that the 2 from the particular supplier go to the first line. Hence,

P(two to first line) =

2.51a. From Table P(A) = .4
  1. From Table P(B) = .37
  2. From Table P(A Ç B) = .10
  3. Using parts a-c: P(A È B) = .40 + .37 -.10 = .67
  4. P(Ac) = .6
  5. [P(A È B)]c = .33
  6. [P(A Ç B)]c = .9
  7. P(A | B) = P(A Ç B)/P(B) = .1/.37
  8. P(B | A) = P(A Ç B)/P(A) = .1/.4
2.65 Let A = Device gets past first inspector and B = device gets past second inspector. We are given P(A) = .1 and P(B | A) = .5. The probability the device gets past both inspectors is:
P(A Ç B) = P(A)P(B | A) = .1 x .5 = .05

2.67 Let A = 666 in CT and B = 666 in PA. We assume that the lottery outcomes are independent. Hence
  1. P(A | B) = P(A) = .001
  2. P(A Ç B) = P(A)P(B) = .001 x 1/8 = .000125
2.74 Let A = innocent person and B = guilty person.
P(A) = .9, P(Ac) = .1, P(B) = .95, P(Bc) = .05.
  1. P(Ac Ç B) = .10 x .95 = .095
  2. P(A Ç B) = .9 x .95 = .855
  3. P(Ac Ç Bc) = .1 x .05 = .005
  4. 1 - P(A Ç Bc) = 1 - .9 x .05 = .955
2.83 D = Has disease and T = Test shows disease. Hence:
P(D) = 0.01, P(T|D) = 0.9, P(Tc|Dc) = 0.9, P(T|Dc) = 0.1

P(Disease|Test shows disease) = P(D|T) =

=

2.85 P = Positive Response, M = Male respondent, F = Female respondent. Hence:
P(M) = , P(F) = 3/4, P(P | F) = .7, P(P | M) = .4. Therefore

P(M | Pc) =

= = .4

2.90 F = Failure to learn and A and B are the two methods. We are given:
P(A) = 0.3, P(B) = 0.7, P(F | A) = 0.2, P(F | B) = 0.1

P(A | F) =

= * = 14/17