2000 Prob-Stat I Midterm Answers Page

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45-733 PROBABILITY AND STATISTICS I Midterm Examination Answers

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

(10 Points)

1. Suppose we have the bivariate discrete probability distribution


                           æ c(4y - 2x)  x = 0, 1, 2
                  f(x,y) = ç             y = 1, 2
                           è      0 otherwise

Find c.


                              y
                           1    2
                         ----------
                       0 | 4    8 | 12 
                         |        |
                    x  1 | 2    6 |  8 
                         |        |
                       2 | 0    4 |  4 
                         |        |
                         ----------
                           6   18 | 24

Therefore, c = 1/24

Find f₁(x) and f₂(y)


                           æ 12/24  x = 0
                           ç 
                           ç  8/24  x = 1      
                   f₁(x) = ç  
                           ç  4/24  x = 2
                           ç  
                           è   0 otherwise

                           æ  6/24  y = 1
                           ç 
                   f₂(y) = ç 18/24  y = 2
                           ç 
                           è   0 otherwise

Are X, Y independent?

NO. Clearly f(x,y) ¹ f₁(x)f₂(y)

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

(10 Points)

2. You draw 5 cards randomly without replacement from a deck of 52 playing cards. What is the probability that there are 2 cards from one suit and 1 card each from three suits (e.g., ©© § ¨ ª )?


                         æ4öæ13öæ13öæ13öæ13ö
                         ç ÷ç  ÷ç  ÷ç  ÷ç  ÷
                         è1øè 2øè 1øè 1øè 1ø
                 P(E) = -------------------
                               æ52ö
                               ç  ÷
                               è 5ø

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

(10 Points)

3. Suppose we have the continuous probability distribution:


                           æ c/x⁴  1 < x < 2
                    f(x) = ç            
                           è   0 otherwise

Find c.

cò₁² 1/x⁴dx = c[-1/(3x³)]|₁² = c(7/24)
Hence, c = 24/7

Find VAR(X).
E(x) = 24/7ò₁² x/x⁴dx = 24/7ò₁² 1/x³dx = (24/7)[-1/(2x²)]|₁² = (9/7)
E(x²) = 24/7ò₁² x²/x⁴dx = 24/7ò₁² 1/x²dx = (24/7)[-1/x]|₁² = (12/7)
VAR(X) = 12/7 - (9/7)² = 3/49

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

(10 Points)

4. Suppose we have the bivariate continuous probability function


                           æ c(2x² - y)  0 < x < 1
                  f(x,y) = ç            -1 < y < 0
                           è 0 otherwise

Find c.

cò_-1⁰ ò₀¹ (2x² - y)dxdy = cò_-1⁰ {[2x³/3 - xy]|₀¹}dy = cò_-1⁰ (2/3 - y)dy = c[(2/3)y - y²/2]|_-1⁰ = c(7/6)
Hence, c = 6/7

Find E(X) and E(Y)
E(x) = 6/7ò₀¹ ò_-1⁰ x(2x² - y)dydx = 6/7ò₀¹ [(2x³y - xy²/2)|_-1⁰]dx = 6/7ò₀¹ (2x³ + x/2)dx = (6/7)[2x⁴/4 + x²/4]|₀¹ = (6/7)(3/4) = 9/14
E(y) = 6/7ò₀¹ ò_-1⁰ y(2x² - y)dydx = 6/7ò₀¹ [(2x²y²/2 - y³/3)|_-1⁰]dx = -6/7ò₀¹ (x² + 1/3)dx = -(6/7)[x³/3 + x/3]|₀¹ = -(6/7)(2/3) = -4/7

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

(10 Points)

5. Suppose we have two urns. In the first urn there are 12 Red and 18 Black balls. In the second urn there are 15 Red and 15 Black balls. One ball is drawn randomly without replacement from each urn. Let Y equal the number of red balls drawn from the two urns.

What is the probability distribution of Y.



                           æ (18/30)(15/30) =                   9/30  y = 0
                           ç 
                           ç (12/30)(15/30) + (18/30)(15/30) = 15/30  y = 1 
                    f(y) = ç
                           ç (12/30)(15/30) =                   6/30  y = 2
                           ç 
                           è   0 otherwise

What is VAR(Y).

E(X) = å_i=1,3 x_if(x_i) = 0*(9/30) + 1*(15/30) + 2*(6/30) = 27/30
E(X²) = å_i=1,3 x_i²f(x_i) = 0*(9/30) + 1*(15/30) + 4*(6/30) = 39/30
VAR(Y) = 39/30 - (27/30)²

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

(10 Points)

6. Suppose we have a discrete bivariate probability distribution


                           æ c(2x + y²)  x = 1, 2, 3
                  f(x,y) = ç             y = 1, 2, 3
                           è      0 otherwise

Find c.


                               y
                           1   2   3
                         ------------
                       1 | 3   6  11 |20 
                         |           |
                    x  2 | 5   8  13 |26 
                         |           |
                       3 | 7  10  15 |32
                         |           |
                         ---------------
                          15  24  39 |78

Hence, c = 1/78

Find P(X > 1 | Y £ 2)

P(X > 1 | Y £ 2) = P(X > 1 Ç Y £ 2)/ P( Y £ 2) = (30/78)/(39/78) = 30/39

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

(10 Points)

7. An urn contains 6 red, 4 white, and 5 blue balls. Three balls are randomly drawn without replacement from the urn. Find the probability that all 3 of the removed balls are blue if it is known that at least 1 of the removed balls is blue.

P(3 are Blue | At Least 1 is Blue) = P(X = 3 | X ³ 1 ) = P(X = 3 Ç X ³ 1 )/ P(X ³ 1 ) =
P(X = 3)/P(X ³ 1 )


                             æ5ö
                             ç ÷
                             è3ø
                 P(X = 3) = -----
                             æ15ö
                             ç  ÷
                             è 3ø

                                     æ10ö
                                     ç  ÷
                                     è 3ø
      P(X ³ 1) = 1 - P(X = 0) = 1 - -----
                                     æ15ö
                                     ç  ÷
                                     è 3ø

                              æ5ö
                              ç ÷
                              è3ø
      P(X = 3)/P(X ³ 1 ) = ----------
                           æ15ö   æ10ö
                           ç  ÷ - ç  ÷
                           è 3ø   è 3ø

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

(10 Points)

8. A test for a particular form of cancer is 99.9% accurate in that if the person being tested has the disease the test will detect it with probability .999. If the person does not have the disease the test will report that the person has the disease with probability .6. The proportion of the population that has the disease is .01. If a person is chosen randomly from the population and the test indicates that she has the disease, what is the probability that in fact she does have the disease.

Let A = "positive result" and D = "Disease"

P(A|D) = 0.999
P(A^c|D) = 0.001
P(A|D^c) = 0.6
P(A^c|D^c) = 0.4
P(D) = 0.01
P(D^c) = 0.99

Hence, P(D|A) = [P(D)P(A|D)]/[P(D)P(A|D) + P(D^c)P(A|D^c)] =
[.01*.999]/[.01*.999 + .99*.6]

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

(10 Points)

9. Suppose we have the bivariate continuous probability distribution


                           æ c(x² + y)  0 < x < 1
                  f(x,y) = ç            1 < y < 2
                           è 0 otherwise

Find c.

cò₁² ò₀¹ (x² + y)dxdy = cò₁² {[x³/3 + xy]|₀¹}dy = cò₁² (1/3 + y)dy = c[y/3 + y²/2]|₁² = c(11/6)
Hence, c = 6/11

Are X and Y Independent?

f₁(x) = 6/11ò₁² (x² + y)dy = 6/11(x² + 3/2)
f₂(y) = 6/11ò₀¹ (x² + y)dx = 6/11(1/3 + y)

Clearly, X and Y are not independent.

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

(10 Points)

10. Suppose we have the probability distribution



                           æ .10 x = 0
                           ç
                           ç .25 x = 2
                           ç
                    f(x) = ç .35 x = 4
                           ç
                           ç .30 x = 8
                           ç
                           è  0 otherwise

Find F(X)


                         æ  0   x < 0
                         ç
                         ç .10  0 £ x < 2
                         ç
                  F(x) = ç .35  2 £ x < 4
                         ç
                         ç .70  4 £ x < 8
                         ç
                         è  1   x ³ 8

What is P(3 < X < 9).

.65