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45-733 PROBABILITY AND STATISTICS I Practice Midterm


Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

1. Consider the following portion of an electric circuit with nine relays. Current will flow from point A to point B if there is at least one complete path when the relays are activated. The relays are turned on simultaneously and their failures are independent of one another. The relays are identical in every respect and the probability that a relay fails is 0.1. What is the probability that current will flow from A to B when the relays are activated? 

                          |--2---3--|
                    |--1--|         |---|
                    |     |--4---5--|   |
                A---|                   |---B
                    |     |-6-| |-7-|   |    
                    |-----|   |-|   |---|
                          |-8-| |-9-|
P[Upper Path Open] = .9*(.81 + .81 - .812) = .86751
P[Lower Path Open] = (.9 + .9 - .81)*(.9 + .9 - .81) = .9801
P[At Least One Path Open] = .86751 + .9801 - .86751*.9801 = .9974

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

2. Suppose we have the continuous bivariate probability distribution: 

                           æ 4(2 - xy)/7  0 < x < 1
                  f(x,y) = ç             0 < y < 1
                           è   0 otherwise
  1. Find f1(x) and f2(y)

    f1(x) = 4/7ò01 (2 - xy)dy = (4/7)(2y - xy2/2)|01 = (4/7)(2 - x/2)
                           æ 4(2 - x/2)/7  0 < x < 1
               f1(x) = ç       
                       è   0 otherwise
    By symmetry 
                           æ 4(2 - y/2)/7  0 < y < 1
               f2(y) = ç       
                       è   0 otherwise
  2. Are X, Y independent?

    No

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

3. You draw 6 cards randomly without replacement from a deck of 52 playing cards. What is the probability that you get a sequence (as defined in class) of denominations?


                         æ8öæ4ö6
                         ç ÷ç ÷
                         è1øè1ø
                 P(E) = -------
                          æ52ö
                          ç  ÷
                          è 6ø

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

4. Suppose we have a continuous probability distribution

                         æ (x3 + 2)/8   0 < x < 2
                  f(x) = ç        
                         è   0 otherwise
  1. Find F(x).

    F(x) = 1/8ò0x (t3 + 2)dt = (1/8)(t4/4 + 2t)|0x = (1/8)(x4/4 + 2x)
                             æ   0               x < 0
                  F(x) = ç (1/8)(x4/4 + 2x)  0 £ x < 2  
                         è   1               x ³ 2
  2. Find P(1/2 < X < 3/2)

    (1/8)[(81/64 + 3) - (1/64 + 1)] = 13/32

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

5. Suppose we have the discrete bivariate probability distribution 

                           æ c(x2y)  x=1, 2, 3
                  f(x,y) = ç         y=1, 2, 3
                           è 0 otherwise
  1. Find c

                                   y
                           1   2   3
                         ------------
                       1 | 1   2   3 | 6 
                         |           |
                    x  2 | 4   8  12 |24 
                         |           |
                       3 | 9  18  27 |54
                         |           |
                         ---------------
                          14  28  42 |84

    Hence, c = 1/84
  2. Find E(X) and E(Y).

    E(X) = (1/84)(1*6 + 2*24 + 3*54) = 216/84
    E(Y) = (1/84)(1*14 + 2*28 + 3*42) = 196/84
  3. Are X, Y Independent.

    Yes. f(x,y) = f1(x)f2(y) because 
                           æ  x2/14  x = 1, 2, 3
               f1(x) = ç       
                       è   0 otherwise

                       æ  y/6    y = 1, 2, 3
               f2(y) = ç       
                       è   0 otherwise

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

6. We have 3 urns. One contains 3 red, 4 white, and 3 blue balls. A second urn contains 5 red, 2 white, and 3 blue balls. The third urn contains 2 red, 2 white, and 6 blue balls. We randomly draw one ball from each urn. What is the probability that the three randomly chosen balls are not all the same color?

P[Not the same color] = 1 - P[All 3 balls are the same color] =
1 - [(3/10)*(5/10)*(2/10) + (4/10)*(2/10)*(2/10) + (3/10)*(3/10)*(6/10)] = 1 - 1/10 = .9

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

7. Suppose we own a website and we keep track of the number of customer hits per second. Let X be the number of customer hits per second. Over a long period of time we find that: 

                         æ .1  x = 10
                         ç
                         ç .2  x = 11
                         ç
                         ç .3  x = 12
                         ç
                  f(x) = ç .2  x = 13
                         ç
                         ç .1  x = 14
                         ç
                         ç .1  x = 15
                         ç
                         è  0 otherwise
Suppose our profit per hit is Y = 0.01X - .03. What is the mean and variance of Y?

E(X) = åi=1,6 xif(xi) = 10*.1 + 11*.2 + 12*.3 + 13*.2 + 14*.1 + 15*.1 = 12.3
E(X2) = åi=1,6 xi2f(xi) = 100*.1 + 121*.2 + 144*.3 + 169*.2 + 196*.1 + 225*.1 = 153.3
VAR(X) = 153.3 - 12.32 = 2.01

E(Y) = .01E(X) - .03 = .01*12.3 - .03 = .093
VAR(Y) = VAR(.01X - .03) = .0001VAR(X) = .0001*2.01 = .0002

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(15 Points)

8. Suppose in a certain state that 55% of the eligible voters are Republicans, 40% are Democrats, and 5% are Reform party members. In the last election 80% of the Republicans voted, 75% of the Democrats voted, and 35% of the Reform party members voted. A person is chosen at random from this population and we learn that she voted in the last election. What is the probability that she is a Democrat?

Let R = "Republican", D = "Democrat", J = "Reform", V = "Voted"

P(R) = .55
P(D) = .40
P(J) = .05
P(V|R) = .80
P(V|D) = .75
P(V|J) = .35

Hence, P(D|V) = [P(D)P(V|D)]/[P(D)P(V|D) + P(R)P(V|R) + P(J)P(V|J)] =
[.40*.75]/[.40*.75 + .55*.80 + .05*.35] = .396

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

9. We roll a die three times. What is the probability that the sum of the number of dots for the three rolls is greater than 15?

P[Sum > 15] = P[Sum = 16] + P[Sum = 17] + P[Sum = 18]
P[Sum = 16] = P[5+5+6] + P[5+6+5] + P[6+5+5] + P[6+6+4] + P[6+4+6] + P[4+6+6] = 6/63 = 6/216
P[Sum = 17] = P[5+6+6] + P[6+5+6] + P[6+6+5] = 3/63 = 3/216
P[Sum = 18] = 1/216
Hence, P[Sum > 15] = 10/216

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Midterm
Keith Poole

(10 Points)

10. Suppose that the student population at GSIA is 70% male and 30% female and that 40% of the males go to tractor pulls and that 10% of the females go to tractor pulls. We randomly draw a student from this population and learn that the person went to the most recent tractor pull at the civic arena. What is the probability that the person is male?

Let M = "Male", F = "Female", T = "Go to Tractor Pull"

P(M) = .70
P(F) = .30
P(T|M) = .40
P(T|F) = .10

Hence, P(M|T) = [P(M)P(T|M)]/[P(M)P(T|M) + P(F)P(T|F)] =
[.70*.40]/[.70*.40 + .30*.10] = 28/31