Exam 1 - Math Solutions

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1 Exam 1 - Math Solutions Spring Without actually expanding, find the coefficient of x y 2 z 3 in the expansion of (2x y z) 6. (A) 120 (B) 60 (C) 30 (D) 20 (E) 10 (F) 10 (G) 20 (H) 30 (I) 60 (J) 120 The monomial we want in the multinomial expansion is ( 6 1,2,3 Therefore, the coefficient is 120. ) (2x)( y) 2 ( z) 3 = 2 6! 1!2!3! x y 2 z 3 = 120x y 2 z 3.

2 2. Given the probabilities P(A)= 0.5 P(A B)=0.3 P(A B C )= 0.1 P(B)=0.6 P(A C )= 0.2 P(C )=0.6 P(B C )=0.3 find the values of P(B A), P(B A c ), and P(A c B C ), respectively. (A) 3/5, 2/3, 4/5 (B) 3/5, 4/5, 2/3 (C) 2/3, 2/3, 3/5 (D) 3/5, 3/5, 1/2 (E) 1/2, 4/5, 2/3 (F) 4/5, 3/5, 2/5 (G) 3/5, 1/2, 1/2 (H) 3/5, 3/5, 2/3 (I) 2/5, 2/5, 2/3 (J) 2/5, 2/5, 3/5 For the first two probabilities we have For the third, P(B A)= B A P(A) = 3 5, P(B Ac )= P(B Ac ) P(B) P(B A) P(A c = = ) 1 P(A) = 3 5. P(A c B C )= P(Ac B C ) P(B C ) P(B C ) P(A B C ) = P(B C ) = =

3 3. The accuracy of a medical diagnostic test, in which a positive result indicates the presence of a disease, is often stated in terms of its sensitivity, the proportion of diseased people who test positive or P(+ Disease), and its specificity, the proportion of people without the disease who test negative or P( No Disease). Suppose that 10% of the population has the disease (called the prevalence rate). A diagnostic test for the disease has 95% sensitivity and 85% specificity. What is the probability that a person who tests positive actually has the disease? (A) 0.35 (B) 0.32 (C) 0.23 (D) 0.15 (E) 0.98 (F) 0.83 (G) 0.75 (H) 0.53 (I) 0.62 (J) 0.41 By Bayes theorem we have P(+ D)P(D) P(D +)= P(+ D)P(D)+P(+ D c )P(D c ) = P(+ D)P(D) P(+ D)P(D)+(1 P( D c ))(1 P(D)) = (1 0.85) (1 0.10) =

4 4. Two fair dice are rolled. Let X be the absolute difference between the outcomes of the dice. What are, respectively, P(0 X < 3) and P(X = 3)? (A) 2/3 and 1/6 (B) 1/3 and 2/3 (C) 1/2 and 1/2 (D) 1/3 and 1/6 (E) 2/3 and 1/3 (F) 1/4 and 2/3 (G) 1/4 and 3/5 (H) 3/5 and 1/3 (I) 2/3 and 1/4 (J) 1/4 and 1/6 The possible values that X can attain are 0,1,2,3,4,5. By simple counting, the number of elementary outcomes (i, j ), where i, j {1,2,...,6}, for each value of X (second line below), and the corresponding probability (third line) is: Then, x # P(0 X < 3)=P(X = 0)+P(X = 1)+P(X = 2)= p = = 2 3 and P(X = 3)=

5 5. A random variable X has the following p.d.f: The mean and variance of X are, respectively (A) 2.50, 0.70 (B) 2.50, 0.75 (C) 1.00, 0.85 (D) 1.50, 0.50 (E) 1.50, 0.75 (F) 1.50, 0.70 (G) 2.00, 0.50 (H) 1.55, 0.71 (I) 1.25, 0.71 (J) 1.50, 0.71 The mean is µ= 1 x f (x)d x= 3x 4 if x 1 f (x)= 0 if x< 1 1 [ 3x 3 d x= 3 ] 2x 2 = = The variance can be calculated using the identity σ 2 = E ( X 2) µ 2. First note: 1 x 2 f (x)d x = 1 [ 3x 2 d x= 3 ] = 3. x 1 Therefore, ( ) 3 2 σ 2 = 3 = =

6 6. (Continuation of the previous problem.) Assume the same information given in problem 5. Then the median of X is (A) 2.26 (B) 1.22 (C) 1.26 (D) 1.49 (E) 1.30 (F) 2.27 (G) 1.10 (H) 1.32 (I) 2.23 (J) 1.31 We first need to obtain the cumulative distribution function. x F (x)= f (s)d s= 1 x 1 3s 4 d s= [ s 3] x 1 = 1 x 3. The median is the value x such that 1 x 3 = 0.5. Solving for x we get x = 2 1/3 =

7 7. In a state lottery game you must match all 6 numbers drawn at random from 1 to 54 without replacement to win the grand prize, 5 out of 6 numbers to win the second prize, and 4 out of 6 numbers to win the third prize. The order of the numbers is irrelevant. Find the probability of winning some prize (first, second, or third). (You may use ( 54) 6 = ) (A) (B) (C) (D) (E) (F) (G) (H) (I) (J) The total number of ways to choose 6 numbers out of 54 is ( 54) 6. Now divide the set of 54 numbers into S w (the winning numbers) and its complement S c w. Then #S w = 6 and #S c w = 48. The number of ways to draw without replacement k numbers from S w and 6 k from the complement is ( 6) ( k 48 6 k). Therefore, the probability we want is ( 6 ( p = 6) 48 ) ( 6 ( 0 ) + 5) 48 ) ( 6 ( 1 ) + 4) 48 ) 2 ) = ( 54 6 ( 54 6 ( 54 6 = =

8 8. A card is drawn successively at random and with replacement from a pack of 52 cards until an ace or a face card is drawn. What is the probability that a face card is drawn before an ace? (A) 0.50 (B) 0.85 (C) 0.90 (D) 0.70 (E) 0.80 (F) 0.55 (G) 0.75 (H) 0.74 (I) 0.72 (J) 0.66 Let E represent the event that the card drawn when the game stops (i.e., when either an ace or a face card is drawn) is a face card. Let A, F, and N represent the events that the first card drawn is, respectively, and ace, a face card, or neither. We want to know the probability P(E). This probability can be written as P(E)=P(E A)P(A)+P(E F )P(F )+P(E N )P(N ). Now P(A)=4/52, P(F )=12/52, P(N )=36/52, P(E A)=0, P(E F )=1, and P(E N )=P(E). Then Solving for P(E), we obtain P(E)=3/4=0.75. P(E)= P(E)

9 9. The two graphs below represent the cumulative distribution functions of two random variables X and Y. What are the probabilities of X = 0.8 and Y = 0.8, respectively? (A) 0.8, 0.8 (B) 0.8, 0.4 (C) 0.8, 0.6 (D) 0.8, 0.2 (E) 0, 0.8 (F) 0, 0.6 (G) 0, 0.2 (H) 0, 0.4 (I) 0.4, 0.8 (J) 0.4, 0.4 From the graph it is clear that X is a continuous random variable. Therefore, P(X = 0.8) = 0. The probability of Y = 0.8 is the size of the jump discontinuity at 0.8, so P(Y = 0.8) = =

10 10. (Continuation of the previous problem.) Assume the same information given in problem 9. What are the probabilities of the intervals 0.2 x < 0.8 and 0.2 < y < 0.8, respectively? (A) 0.6, 0.6 (B) 0.4, 0.6 (C) 0.8, 0.6 (D) 0.6, 0.4 (E) 0.4, 0.4 (F) 0.8, 0.4 (G) 0.6, 0 (H) 0.2, 0 (I) 0.4 and 0 (J) 0.8, 0 We have for X : For Y, P(0.2 X 0.8)= F (0.8) F(0.2)= =0.4. P(0.2<Y < 0.8)= lim a 0+ F (0.8 a) F(2)= = 0. 10

11 11. Let X 0, X 1, and X 2 be three uncorrelated random variables with variances 2,2,7, respectively. What is the correlation between Y 1 = X 0 X 1 and Y 2 = X 0 X 2? (A) 0.71 (B) 0.73 (C) 0.74 (D) 0.75 (E) 0.25 (F) 0.31 (G) 0.50 (H) 0.35 (I) 0.33 (J) 0.67 Since the X i are uncorrelated, Cov(Y 1,Y 2 )=Cov(X 0 X 1, X 0 X 2 )=Var(X 0 ) Cov(X 0, X 1 ) Cov(X 0, X 2 )+Cov(X 1, X 2 )=Var(X 0 ). Similarly, Therefore, Var(Y 1 )=Var(X 0 )+Var(X 1 ), and Var(Y 2 )=Var(X 0 )+Var(X 2 ). Corr(Y 1,Y 2 )= Var(X 0 ) Var(X0 )+Var(X 1 ) Var(X 0 )+Var(X 2 ) = =

12 12. Consider an insurance agency with customers who have both auto and homeowners policies. Let X and Y be discrete random variables, where X is the deductible amount on the auto policy and Y is the deductible amount on the homeowner policy. A summary of these policies is shown below. X Y Totals $0 $100 $200 $ $ The conditional probability of Y = $100 given that X = $250 is (A) 0.30 (B) 0.20 (C) 0.25 (D) 0.32 (E) 0.60 (F) 0.63 (G) 0.75 (H) 0.15 (I) 0.27 (J) 0.33 This information can be read from the second entry of the second row of the table: P(Y = $100 X = $250)= =

13 13. (Continuation of the previous problem.) Assume the same information given in problem 12. Then the mean values µ X and µ Y, and variances σ 2 X and σ2 Y of X and Y are: (A) µ X = 170, µ Y = 165, σ 2 X = 5620, σ2 Y = 6855 (B) µ X = 175, µ Y = 125, σ 2 X = 5625, σ2 Y = 6875 (C) µ X = 180, µ Y = 265, σ 2 X = 5620, σ2 Y = 6855 (D) µ X = 179, µ Y = 160, σ 2 X = 5020, σ2 Y = 2855 (E) µ X = 175, µ Y = 125, σ 2 X = 7625, σ2 Y = 7875 (F) µ X = 270, µ Y = 365, σ 2 X = 5699, σ2 Y = 6855 (G) µ X = 175, µ Y = 125, σ 2 X = 5925, σ2 Y = 7870 (H) µ X = 170, µ Y = 265, σ 2 X = 4020, σ2 Y = 6555 (I) µ X = 175, µ Y = 125, σ 2 X = 5655, σ2 Y = 6890 (J) µ X = 170, µ Y = 165, σ 2 X = 9620, σ2 Y = 6809 It is convenient to write the matrix of probabilities: $0 $100 $200 $ $ Then µ X = 0.50 $ $250 = $175 µ Y = 0.25 $ $ $200 = $125 σ 2 X = 0.50 ($100) ($250) 2 ($175) 2 = 5625 σ 2 Y = 0.25 ($0) ($100) ($200) 2 ($125) 2 =

14 14. (Continuation of the previous problem.) Assume the same information given in problem 12. The covariance and correlation of X and Y are, respectively (A) 4750, (B) 3673, (C) 1253, (D) 0.302, 1875 (E) 3209, (F) 1953, (G) 2794, (H) 1875, (I) 0.391, 3673 (J) 0.223, 1253 The covariance of X and Y is Cov(X,Y )=0.20 $100 $ $100 $ $100 $ $250 $ $250 $ $250 $200 µ X µ Y = = Now the correlation follows: Corr(X,Y )= Cov(X,Y ) 1875 = = σ x σ Y

Homework 4 Solution, due July 23

Homework 4 Solution, due July 23 Random Variables Problem 1. Let X be the random number on a die: from 1 to. (i) What is the distribution of X? (ii) Calculate EX. (iii) Calculate EX 2. (iv) Calculate Var