STAT/MA 416 Midterm Exam 1 Tuesday, September 18, Circle the section you are enrolled in:

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1 STAT/MA 416 Midterm Exam 1 Tuesday, September 18, 2007 Name Purdue student ID (10 digits) Circle the section you are enrolled in: STAT/MA STAT/MA :00 AM 10:15 AM 3:00 PM 4:15 PM REC 114 UNIV The testing booklet contains 12 questions. 2. Permitted Texas Instruments calculators: BA-35 BA II Plus BA II Plus Professional TI-30Xa TI-30X IIS (solar) TI-30X IIB (battery) The memory of the calculator should be cleared at the start of the exam, if possible. 3. Mark your answers on the Scantron sheet using a #2 pencil. 4. Make sure to supply answers to all of the questions. There is no penalty for guessing. 5. No partial credit will be given. 6. Show all of your work in the exam booklet. 7. Extra sheets of paper are available from the instructor.

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3 1. Suppose that four urns each contain two balls. Urns 1 and 2 each have one black ball and one white ball. Urn 3 has two black balls. Urn 4 has two white balls. An urn is selected at random. Given that the first ball drawn from this urn is black, what is the probability that the other ball in this urn is also black? A.) 1/8 B.) 1/6 C.) 1/4 D.) 1/3 E.) 1/2 Answer. Write E i for the event that Urn i is selected. Write B 1 for the event that the first ball drawn is black, and write B 2 for the event that the second ball drawn is black. Then the desired probability is P (B 2 B 1 ) P (B 1 B 2 ) P (B 1 ) which can be rewritten as P (B 1 B 2 E 1 )P (E 1 ) + P (B 1 B 2 E 2 )P (E 2 ) + P (B 1 B 2 E 3 )P (E 3 ) + P (B 1 B 2 E 4 )P (E 4 ) P (B 1 E 1 )P (E 1 ) + P (B 1 E 2 )P (E 2 ) + P (B 1 E 3 )P (E 3 ) + P (B 1 E 4 )P (E 4 ) which simplifies to So the correct answer is E, namely, 1/2. (0)(1/4) + (0)(1/4) + (1)(1/4) + (0)(1/4) (1/2)(1/4) + (1/2)(1/4) + (1)(1/4) + (0)(1/4) 1 2 A simple way to look at the problem is this: There are 4 black balls; each of these four is equally likely to be picked on the first draw. Exactly two of these four balls are from Urn 3, so choosing exactly two of these four balls will make the second ball also be black. So the desired conditional probability is 2/4 1/2. 1

4 2. A box has 10 balls, 6 of which are black and 4 of which are white. Three balls are removed from the box, their color unnoted. Find the probability that a fourth ball removed from the box is white. A.) 1/10 B.) 1/4 C.) 1/3 D.) 2/5 E.) 2/3 Answer. Any of the 10 balls is equally likely to be drawn on the fourth time. So the probability that the fourth ball is black is 4/10 2/5. So the correct answer is D, namely, 2/5. 2

5 3. In answering a question on a multiple-choice test, a student either knows the answer or guesses. Let p.70 be the probability that the student knows the answer and 1 p.30 the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probabilty 1/5, since 5 is the number of multiple-choice alternatives. What is the conditional probability that a student knew the answer to a question, given that he or she answered it correctly? A.).43 B.).70 C.).74 D.).90 E.).92 Answer. Write E for the event that the student knew the answer to a question, and write F for the event that he or she answered it correctly. So the desired probability is P (E F ) P (E F ) P (F ) P (F E)P (E) P (F E)P (E) + P (F E c )P (E c ) (1)(.70) (1)(.70) + (1/5)(.30).92 So the correct answer is E, namely,.92. 3

6 4. Urn I contains 2 black balls and 2 white balls; urn II contains 1 black ball and 2 white balls; urn III contains 3 black balls and 1 white ball. One urn is selected at random, and then one ball is drawn from it. Given that a black ball is drawn, what is the probability that urn I was selected? A.) 2/11 B.) 6/19 C.) 2/6 D.) 19/36 E.) 6/11 Answer. Write E i for the event that Urn i was selected, and write F for the event that a black ball was drawn. Then the desired probability is P (E 1 F ) P (E 1 F ) P (F ) P (F E 1 )P (E 1 ) P (F E 1 )P (E 1 ) + P (F E 2 )P (E 2 ) + P (F E 3 )P (E 3 ) (2/4)(1/3) (2/4)(1/3) + (1/3)(1/3) + (3/4)(1/3) 6/19 So the correct answer is B, namely, 6/19. 4

7 5. In general, after a group makes a reservation at a restaurant, they do not show up 20% of the time. If a restaurant has 10 tables and takes 12 reservations, what is the probability that it will be able to accommodate everyone? A.).73 B.).77 C.).83 D.).85 E.).91 Answer. The restaurant will be able to accommodate everyone if 10 or fewer groups show up, i.e., neither 11 nor 12 groups show up. So the desired probability is ( ) ( ) (.80) 11 (.20) 1 (.80) 12 (.20) So the correct answer is A, namely,.73. 5

8 6. If it is assumed that all ( 52 5 ) poker hands are equally likely, what is the probability of being dealt a full house (i.e., one pair and one triple of cards with equal face values)? (Reminder: A deck has 52 cards in 4 suits with 13 values: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.) A.) B.) C.) D.) E.) Answer. There are ( ) ways to choose the value for the triple of cards; for each such choice, there are ( ) remaining ways to choose the value for the pair of cards. There are ( ( 4 3) ways to choose a triple with a given value, and there are 4 2) ways to choose a pair with a given value. There are ( ) 52 5 hands altogether. So the desired probability is exactly ( 13 )( 12 )( 4 )( ) 2) ( 52 5 So the correct answer is D, namely,

9 7. From a group of 10 women and 10 men a committee consisting of 3 women and 3 men is to be formed. How many different committees are possible if 1 woman and 1 man refuse to serve together? A.) 64 B.) 1296 C.) 3456 D.) E.) Answer. There are ( ) ( 10 3 ways to pick 3 women, and for each such choice, there are 10 ) 3 ways to pick 3 men. So there are ( )( ) committees possible. Now we consider the number of committees that are forbidden, i.e., that contain the feuding man and woman. There are two other women needed for each such committee and two other men needed for each such committee, so that there are ( 9 9 2)( 2) 1296 forbidden committees. So the total number of allowed committees is So the correct answer is D, namely,

10 8. Ten toys are to be given to five children A, B, C, D, E. How many ways can the toys be assigned if the toys are indistinguishable (i.e., look identical)? What if the toys are distinguishable (i.e., look different)? A.) 252 ways with indistinguishable toys; ways with distinguishable toys B.) 3003 ways with indistinguishable toys; ways with distinguishable toys C.) 1001 ways with indistinguishable toys; ways with distinguishable toys D.) 1001 ways with indistinguishable toys; ways with distinguishable toys E.) 3003 ways with indistinguishable toys; ways with distinguishable toys Answer. The number of assignments of indistinguishable (i.e., identical-looking) toys is the ( number of solutions of x 1 + x 2 + x 3 + x 4 + x 5 10 with x i 0 for all i, which is ) ( ) The number of ways of assigning distinguishable (i.e., different-looking) toys is 5 ways per toy, which is altogether. So the correct answer is D, namely, 1001 ways with indistinguishable toys; ways with distinguishable toys. 8

11 9. Twenty-five indistinguishable (i.e., identical-looking) balls are placed in three urns, labelled A, B, C. How many such assignments result in at least one ball in urn A, at least two balls in urn B, and at least three balls in urn C? A.) 171 B.) 210 C.) 231 D.) 300 E.) 2300 Answer. The number of assignments is the number of solutions of x 1 + x 2 + x 3 25 with x 1 1 and x 2 2 and x 3 3. We can define y 1 x and y 2 x and y 3 x 3 1 0, so that x 1 +x 2 +x 3 25 becomes (x 1 1)+(x 2 2)+(x 3 3) , or equivalently, y 1 + y 2 + y 3 19 with y i 0 for all i. There are ( ) ( ) solutions to this equation (as discussed in Proposition 6.2). So the correct answer is B, namely, 210. Another possible method of solution is to define y 1 x 1 1 and y 2 x and y 3 x 3 2 1, so that x 1 + x 2 + x 3 25 becomes x 1 + (x 2 1) + (x 3 2) , or equivalently, y 1 + y 2 + y 3 22 with y i 1 for all i. There are ( ) ( ) solutions to this equation (as discussed in Proposition 6.1). 9

12 10. An urn contains 7 balls, of which 2 are white. Two players A, B successively draw from the urn, A first, then B, then A, then B, then A, and so on. The winner is the first one to draw a white ball. What is the probability B wins if each ball is replaced after it is drawn? What is the probability B wins if the withdrawn balls are not replaced? A.) with replacement, P (B wins) 2/9; without replacement, P (B wins) 3/7 B.) with replacement, P (B wins) 5/12; without replacement, P (B wins) 3/7 C.) with replacement, P (B wins) 5/12; without replacement, P (B wins) 8/21 D.) with replacement, P (B wins) 2/9; without replacement, P (B wins) 5/21 E.) with replacement, P (B wins) 5/7; without replacement, P (B wins) 8/21 Answer. With replacement, there are two possible methods of calculating the desired probability. First, we just directly write the sum of the probabilities that B wins on his first, second, third, etc., try: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P (B wins) (( ) ( )) n ( ) ( ) n0 ( ) ( ) ( ( 5 5 ) 7 5/12 Another possible method of solution when the balls are replaced is the following: The probability that a round (i.e., A then B) will complete the game is ( ( 2 2 ( + 2 ( 5 ( + 5 ( 2, and of these three possibilities, exactly one corresponds to the probability that B is the one of complete the game, namely ( ( 5 2. So the probability that B wins is ( 5 ) ( 2 ( 7 2 ) ( 2 ) ( ) ( 5 ) ( ) ( 2 5/12 7 When the balls are withdrawn without replacement, we just calculate directly the probability that B wins, namely: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) / So the correct answer is B, namely, with replacement, P (B wins) 5/12; replacement, P (B wins) 3/7. without 10

13 11. Suppose we have 10 coins such that if the nth coin is flipped, heads will appear with probability 1/n, for n 1, 2,..., 10. When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the sixth coin? A.).0030 B.).0182 C.).0569 D.).1091 E.).3414 Answer. Write E i for the event that the ith coin is selected, and write F for the event that the selected coin shows heads. Then the desired probability is P (E 6 F ) P (E 6 F ) P (F ) P (F E 6 )P (E 6 ) P (F E 1 )P (E 1 ) + P (F E 2 )P (E 2 ) + P (F E 3 )P (E 3 ) + + P (F E 10 )P (E 10 ) (1/6)(1/10) (1/1)(1/10) + (1/2)(1/10) + (1/3)(1/10) + + (1/10)(1/10).0569 So the correct answer is C, namely,

14 12. Die A has four red and two white faces, whereas die B has two red and four white faces. A fair coin is flipped once at the start of a game. If the coin lands heads, then die A is used throughout the game. If the coin lands tails, then die B is used throughout the game. If the first n throws of the die are red, what is the probability that die A is being used? 2 A.) n B.) (1/2) n C.) (2/3) n 2 D.) n 3 E.) n 3 n +1 2 n +1 3 n +1 Answer. Write H for the event that the coin shows heads. Write E 1 for the event that die A is used, and write E 2 for the event that die B is used. Write F for the event that the first n throws of the die are red. Then the desired probability is P (E 1 F ) P (E 1 F ) P (F ) P (F E 1 )P (E 1 ) P (F E 1 )P (E 1 ) + P (F E 2 )P (E 2 ) (4/6) n (1/2) (4/6) n (1/2) + (2/6) n (1/2) (2/3) n (1/2) (2/3) n (1/2) + (1/3) n (1/2) We simplify by multiplying the numerator and denominator by 2 and also by 3 n. So the 2 desired probability is n. 2 n +1 So the correct answer is D, namely, 2 n 2 n