EXAM. Exam #1. Math 3371 First Summer Session June 12, 2001 ANSWERS
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1 EXAM Exam #1 Math 3371 First Summer Session 2001 June 12, 2001 ANSWERS
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3 Give answers that are dollar amounts rounded to the nearest cent. Here are some possibly useful formulas: A = P (1 + rt), A = P (1 + i) n, r e = (1 + r/m) m 1 S = R (1 + i)n 1 1 (1 + i) n, P = R. i i Problem 1. Suppose that you open a savings account with a deposit of $4,000. The account pays 7.5% interest, compounded monthly. If no further deposits are made, how much is in the account after 10 years? This is a compound interest problem, so we use the formula A = P (1+i) n. Here P = $4000 is the principal, the interest rate per period is i = r/m = 0.075/12 = and n = 12(10) is the number of periods. Thus, A = 4000( ) 20. Evaluating this on a calculator and rounding to the nearest cent gives A = $ Problem 2. A savings account at Bank A pays 8% compounded qarterly and a savings account at Bank B pays 7.95% compounded monthly. Find the effective rate in each case. Which account is a better investment? To compare the two accounts, compare the effective rates (which tells you what your money earns in one year). For Bank A, we have an interest rate of r = 0.08 and the number of compounding periods is m = 4. Thus, the effective rate is r e = ( /4) For Bank B we have an interest rate of r = and the number of compounding periods is m = 12, so the effective rate is Thus, we see that r e = ( /12) Bank B offers a better investment. Problem 3. Suppose that $125 is deposited at the end of each month into a savings account that pays 6% interest compounded monthly. How much is in the account after 8 years? 1
4 The problem asks for the future value of an ordinary annuity, so the formula is S = R (1 + i)n 1. i The periodic deposit is R = $125. The interest rate per period is i = r/m = 0.06/12 = and the number of periods is n = 8(12) = 96. Thus, we have S = 125 (1.005) $15, Problem 4. Gina Simone is saving for a VCR. At the end of each month she makes a deposit in a savings account that pays 6% compounded monthly. How much should she deposit each month to have $400 in the account at the end of one year? This is a sinking fund problem we want to find the periodic payment R for an ordinary annuity so that the future value S is a given value, S = 400 in this case. Solving the future value formula for R, we get i R = S (1 + i) n 1. In the present situation, we have S = 400, the amount we want in the account at the end. The interest rate per period is i = 0.06/12 = and the number of periods is n = 12. Thus, R = 400 (1.005) 12 1 $ Problem 5. Suppose that you borrow $150,000 to buy a house on a 30 year mortgage. The interest rate is 7.2% compounded monthly. How much will your monthly payment be? How much interest will you pay over the life of the loan? Use the formula for the present value of an annuity. amount R. Solving for R, we have We want the payment R = P i 1 (1 + i) n. 2
5 In the present situation, P = 150, 000, the amount of the loan. The interest rate per period is i = 0.072/12 = and the number of payments is 30(12) = 360. Thus, the monthly payment will be R = 150, 000 $ ;. 1 (1.006) 360 The interest paid over the life of the loan is the total of all the payments minus the amount of the loan. Thus, Total Interest = ( )(360) 150, 000 = $216, Problem 6. An urn contains 4 balls lettered A, B, C, and D. A hat contains a slip of black paper and a slip of yellow paper. An experiment consists of selecting one ball from the urn and one slip of paper from the hat. Write out the sample space, and then write the indicated events in set notation. A. Event P, the ball is lettered A or C. We can represent an outcome of the experiment by an ordered pair, where the first element tells us which ball was drawn (written A, B, C or D) and the second element tells us what color of paper was drawn from the hat. Let s write y for the yellow slip of paper and b for the black slip of paper. Thus, we can represent the sample space S by the set S = { (A, y), (A, b), (B, y), (B, b), (C, y), (C, b), (D, y), (D, b) }. The event P consists of all the outcomes where ball A or C was drawn, and so is represented by the set of pairs where the first element is A or C. Thus, P = { (A, b), (A, y), (C, b), (C, y) }. B. Event Q, the paper is yellow. The event Q consists of the outcomes where the yellow slip is drawn from the hat, and hence is represented by the set of all pairs where the second element is y. Thus, Q = { (A, y), (B, y), (C, y), (D, y) }. 3
6 30 pts. Problem 7. ccla urn contains 3 red balls, 2 white balls, 4 blue balls and 1 green ball. If a ball is taken from the urn at random, find the probability that it is A. blue; 4/10 B. not red; 7/10 C. white or blue; 6/10 Problem 8. A single card is drawn at random from a standard deck of 52 cards. Find the probability of drawing the following cards A. A 6 or a jack Let S be the event of drawing a 6 and let J be the event of drawing a jack. Since there are 4 cards of each denomination, P (S) = P (J) = 4/52. By the union rule, P (S J) = P (S) + P (J) P (S J). In this case, S J =, since the card can t be both a 6 and a Jack. Thus, P (S J) = P (S) + P (J) 0 = = ;. B. A black card or a 10 Let B be the event that the card is black and let T be the event that the card is a ten. Half the cards are black, so P (B) = 1/2. There are 4 tens in the deck, so P (T ) = 4/52. These events are not mutually exclusive; there are two black tens. Thus, P (B T ) = 2/52. By the union rule, P (B T ) = P (B) + P (T ) P (B T ) = =
7 C. A heart or a club Let H be the event that the card is a heart and let C be the event that the card is a club. There are 13 cards in each suit. Thus, P (H) = P (C) = 13/52 = 1/4. The events H and C are mutually exclusive, so P (H C) = P (H) + P (C) = = 1 2. D. A red card or a face card Let R be the event that the card drawn is red, and let F be the event that it is a face card. Half of the cards in the deck are red, so P (R) = 1/2. There are (4)(3) = 12 face cards in the deck, so P (F ) = 12/52. These events are not mutually exclusive there are 6 red face cards. Thus, P (R F ) = 6/52. By the union rule, P (R F ) = P (R) + P (F ) P (R F ) = = Problem 9. Two cards are drawn at random from a standard deck of 52 cards, without replacement. Find the probabilities of the following results. A. Both cards are aces Let A 1 denote the event that the first card drawn is an ace, and let A 2 be the event that the second card drawn is an ace. The event that both cards are aces is A 1 A 2. By the multiplication rule for intersections of events, P (A 1 A 2 ) = P (A 2 A 1 )P (A 1 ). When drawing the first card, there are 52 cards in the deck, of which 4 are aces. Tkus, P (A 1 ) = 4/52. If an ace is drawn for the first card, 51 cards remain in the deck, of which 3 are aces, so P (A 2 A 1 ) = 3/51. Thus, P (A 1 A 2 ) = =
8 B. At least one card is red Let R 1 be the event that the first card is read and let R 2 be the event that the second card drawn is read. The event we re looking for is is R 1 R 2. This computation is probably best represented by a probability tree (see Figure 1).!"#$% Figure 1: Probability Tree for Problem 9B Here R stands for drawing a red card and B stands for drawing a black card. The probabiites for the second draw are calculated by considering the cards remaining in the deck. For example, if we draw a red card on the first draw, there are 51 cards remaining, of which 25 are red and 26 are black. The probability of getting at least one red card would be the sum of the products along the 3 branches of the tree that have at least one R. Thue, the probability of at least one red is = drawing a red card on the first draw, R 2 stands for the probability of drawing a red card on the second draw, and so forth, the probability of getting at 6
9 least one red card is P (R 1 R 2 ). We have as before. P (R 1 R 2 ) = P (R 1 ) + P (R 2 ) P (R 1 R 2 ) = P (R 1 ) + P (R 2 (R 1 B 1 )) P (R 1 R 2 ) = P (R 1 ) + P (R 2 R 1 ) + P (R 2 B 1 ) P (R 1 R 2 ) = P (R 1 ) + P (R 2 B 1 ) = P (R 1 ) + P (R 2 B 1 )P (B 1 ) = = , C. The second card is red, given that the first is black. This was already calculated in making the diagram in Figure 1. The probability is P (R 2 B 1 ) = D. The second card is not a face card, given that the first card was a face card. Althogh it s not really necesare, we can draw out the whole probability tree, using R to stand for drawing a face card and R for the complementary event of not drawing a face card. See Figure 2, recalling that there are 12 face cards and 40 non-face cards in the deck. The probability of getting a noe-face card on the second draw given that a face card was drawn on the first draw is40/51, since if the first card drawn is face card, there are 51 cards left, of which 40 are non-face cards. Thus, P (R F ) = Problem 10. The table below shows the results of a survey of 400 customers at a fast food restaurant that serves hamburgers and chicken sandwiches. The customers were asked if they were satisfied with the food they ordered. 7
10 !" #$%&# ')(*+, Figure 2: Probability tree for Problem 9D 8
11 Satisfied Not Satisfied Totals Hambuger Chicken Totals Let A represent the event ordered a hamberger and S represent the event statisfied. Find P (S A ). We want to use the formula P (S A ) = P (S A ) P (A. ) Since people who did not order a burger must have ordered a chicken sandwich, P (A ) is the probability that a customer ordered chicken. From the table, we see that there are 400 customers altogether, and 145 ordered chicken, so P (A ) = 145/400 The customers in S A are the customers who ordered chicken and were satisfied. Looking at the instesection of row 1 of the table with column 1, we see that the numver of customers in S A is 125, thus P (S A ) = 125/400. Then we have P (S A ) = 125/ /400 = ;. Problem 11. An urn contains 3 red balls and 6 green balls. Two balls are drawn, one at a time, without replacement. Construct the probability tree and find the probabilities of the following events. A. The second ball is green, given that the first ball is red. If the first ball is red, there are 8 balls remaining in the urn, or which 2 are red and 6 are green, thus P (second green first red) = 6 8 = 3 4. The same reasoning is used to construct the whole probability tree. Let R stand for drawing a red ball and G stand for drawing a green ball, G 1 for drawing a green ball on the first draw, G 2 for drawing a green ball on the second draw, and so forth. The whole probability tree is given in Figure 3 B. The second ball is green. 9
12 * +, -.#, /#0!#" 1 2'3 4 5#6 7#8 $ %'& (#) 9 : ; < B C'D E FD GH Figure 3: Probability tree for Problem 11 10
13 This would be the sum of the products along all the branches ending in G. Thus, P (G 2 = = 2 3 This, of course, amounts to saying P (G 2 ) = P (G 2 R 1 )P (R 1 ) + P (G 2 G 1 )P (G 1 ). C. The first ball is red, given that the second ball is green (Hint: Bayes Theorem). The probability asked for is P (R 1 G 2 ). To find this, we have to apply Bayes Theorem. Bayes Theorem says, in this case, P (R 1 G 2 ) = P (G 2 R 1 )P (R 1 ) P (G 2 R 1 )P (R 1 ) + P (G 2 G 1 )P (G 1 ), or, to put it another way, the product along the branch involving G 2 and R 1, divided by the sum of the products along the branches that end in G. Thus, P (R 1 G 2 ) = =
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