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
- From Table P(B) = .37
- From Table P(A Ç
B) = .10
- Using parts a-c: P(A È
B) = .40 + .37 -.10 = .67
- P(A^{c}) = .6
- [P(A È
B)]^{c} = .33
- [P(A Ç
B)]^{c} = .9
- P(A | B) = P(A Ç
B)/P(B) = .1/.37
- 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
- P(A | B) = P(A) = .001
- 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(A^{c}) = .1, P(B) = .95,
P(B^{c}) = .05.
- P(A^{c} Ç
B) = .10 x .95 = .095
- P(A Ç
B) = .9 x .95 = .855
- P(A^{c} Ç
B^{c}) = .1 x .05 = .005
- 1 - P(A Ç
B^{c}) = 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(T^{c}|D^{c}) = 0.9,
P(T|D^{c}) = 0.1
P(Disease|Test shows disease) = P(D|T) =
=
2.85 P = Positive Response, M = Male respondent,
F = Female respondent. Hence:
P(M) = ¼, P(F) = 3/4, P(P | F) = .7,
P(P | M) = .4. Therefore
P(M | P^{c}) =
=
= .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