ECE 302 Homework Assignment 2 Solutions
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1 ECE 302 Assignment 2 Solutions January 29, ECE 302 Homework Assignment 2 Solutions Note: To obtain credit for an answer, you must provide adequate justification. Also, if it is possible to obtain a numeric answer, you should not only find an expression for the answer, but also solve for the numeric answer. 1. Consider shuffling a deck of cards and drawing one. Suppose that the card is black. (a) What is the probability that the card is the 4 of clubs? P [4 B] = P [4 B] = 1/52 1/2 = 1/26 (b) What is the probability that the card is a face card (J,Q,or K)? P [F B] = P [F B] (c) What is the probability that the card is a spade? P [ B] = P [ B] = 6/52 1/2 = 3/13 = 13/52 1/2 = 1/2 (d) What is the probability that the card is a 4, 5, or 6? P [(4 5 6) B] = P [4 B] + P [5 B] + P [6 B] = 6/52 1/2 = 3/13 where we have used disjointness of sets corresponding to outcomes 4, 5, and Consider shuffling a deck of cards and drawing one. Suppose that the card is a face card. (a) What is the probability that the card is a jack? P [J F ] = P [J F ] = 4/52 12/52 = 1/3
2 ECE 302 Assignment 2 Solutions January 29, (b) What is the probability that the card is red? P [R F ] = P [R F ] = 6/52 12/52 = 1/2 (c) What is the probability that the card is a diamond? P [ F ] = P [ F ] (d) What is the probability that the card is a 4? P [4 F ] = P [4 F ] = 3/52 12/52 = 1/4 = 0/52 12/52 = 0 3. Consider shuffling a deck of cards and drawing one. Suppose that the card is a 4. Draw another card from the remainder of the deck. (a) What is the probability that only one of the cards is a 4? The probability that only one of the two is a four is the probability that the second one drawn from the remaining cards is not a 4. There are 3 remaining 4 s so the probability that we choose something else is (51-3)/51 = 48/51. (b) What is the probability that both of the cards are red? Obviously, the fact that a 4 was drawn does not tell us anything about whether it is red. The probability that two red cards is drawn is the number of ways to choose two red cards over the number of ways to choose 2 cards, i.e. (26)25/(52(51)) = 25/102. (c) What is the probability that at least one of the cards is a diamond? The probability that at least one is a diamond is 1 minus the probability that both are not diamonds. The number of ways to choose two non-diamonds is 39(38). The numbers of ways to choose two cards is 52(51). The probability requested is thus 1-(39(38))/(52(51)) = 1-19/34 = 15/34. (d) What is the probability that the sum of the values of the two cards is 7? If the two cards are to add to seven, the second must be a three. There are 4 threes among the remaining 51 cards so the probability is 4/51 4. Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was found that 16% carried Lyme disease, 10% had HGE, and 10% of those that had [at least] one disease had both. Yates and Goodman
3 ECE 302 Assignment 2 Solutions January 29, (a) Let L designate the event that the tick has Lyme disease and H the event that the tick has HGE. Suppose that P [L H] = x. Give an expression for P [L H] in terms of P [L], P [H], and x. P [L H] = P [L] + P [H] P [L H] = x = 0.26 x (b) What is P [L H L H]? The probability that the tick has both given that the tick has one is given. It is 10%. (c) Now find an expression for P [L H] in terms of known quantities and solve for the probability that the tick has both diseases. 0.1 = P [L H L H] = P [(L H) (L H)] P [L H] = P [L H] P [L H] = 0.26 x x so 0.26 x = 0.1x so 0.26 = 1.1x and thus x = 0.26/1.1 = 26/110 = 13/55. (d) What is the conditional probability that a tick has HGE given that it has Lyme disease? P [H L] = P [HL] P [L] = 0.26 x 0.16 = / = 0.26(0.1) 1.1(0.16) = 26 16(11) = (e) What is the conditional probability that a tick has Lyme disease given that it has HGE? P [L H] = P [LH]/P [H] = (13/55)/0.1 = 13/ Consider rolling a seven-sided die. Let R j signify the event that the roll is j, L j signify the event that the roll less than j, E be the event that the roll is even. (a) What is P [R 2 L 5 ]? (b) What is P [L 5 E]? P [R 2 L 5 ] = P [R 2L 5 ] P [L 5 ] P [L 5 E] = P [L 5 E] = P [R 2] P [L 5 ] = 1/7 4/7 = 1 4 = 2/7 3/7 = 2 3
4 ECE 302 Assignment 2 Solutions January 29, (c) What is the conditional probability that the roll is less than 4 given that it is even? P [L 4 E] = P [L 4 E] = 1/7 3/7 = 1 3 (d) What is the conditional probability that the 3 is rolled given that the roll is less than 4? P [R 3 L 4 ] = P [R 3L 4 ] = P [R 3] P [L 4 ] P [L 4 ] = 1/7 3/7 = Consider rolling an eight-sided die. Let R j signify the event that the roll is j, L j signify the event that the roll less than j, E be the event that the roll is even. (a) What is P [R 2 L 5 ]? (b) What is P [L 5 E]? P [R 2 L 5 ] = P [R 2L 5 ] P [L 5 ] P [L 5 E] = P [L 5 E] = P [R 2] P [L 5 ] = 1/8 4/8 = 1 4 = 2/8 4/8 = 1 2 (c) What is the conditional probability that the roll is less than 4 given that it is even? P [L 4 E] = P [L 4 E] = 1/8 4/7 = 1 4 (d) What is the conditional probability that the 3 is rolled given that the roll is less than 4? P [R 3 L 4 ] = P [R 3L 4 ] = P [R 3] P [L 4 ] P [L 4 ] = 1/8 3/8 = Consider drawing tiles at random from a set of five scrabble tiles {a, b, c, d, e}. (a) Suppose that someone else draws one tile and tells you that it is neither a nor b. What is the probability that it is c? Let A be the event that the tile a is drawn, B the event that the tile b is drawn, etc. Let X be the event that neither a nor b is drawn. Then P [C X] = P [CX] P [X] = 1/5 3/5 = 1 3.
5 ECE 302 Assignment 2 Solutions January 29, (b) Suppose that someone else draws one tile and tells you that it is neither a nor b. What is the probability that it is either c or d? P [C D X] = P [C D] P [X] where we have used the fact that C D =. = 1/5 + 1/5 3/5 (c) Now suppose that they give you the remaining tiles and you draw one. What is the probability that the tile is a? In this case we know that both a and b are among the remaining four tiles so P [A] = 1/4. (d) What is the probability that the tile you drew is c? Now let C1 be the event that the first tile chosen is not c and C 2 be the event that the second tile chosen is c. P [C 2 ] = P [C 2 C 1 ]P [ C 1 ] = (1/4)(2/3) = 1/6. 8. Now suppose the tiles are numbered rather than lettered and the numbers are 1, 2, 3, 4, and 5. Select tiles at random, one-by-one. Let E j be the event that the jth tile drawn is even. (a) What is P [E 3 E 1 ]? We ll need the probability that both are even. There are 12 possible sequences in which the first and third are both even. There are 5! possible sequences, thus P [E 3 E 1 ] = 12/120 = 1/10. Now P [E 3 E 1 ] = P [E 3E 1 ] P [E 1 ] = 1/10 2/5 = 1 4. (b) What is the probability that the second tile drawn is odd given that the first is? Let O j be the event that the jth tile drawn is odd. Then P [ ] = P [O 2O 1 ] P [O 1 ] = (3/5)(2/4) 3/5 = 2 3 = 1 2. (c) What is the conditional probability that the first two tiles drawn are odd given that third one was? P [(O 1 O 2 ) O 3 ] = P [O 1 O 2 O 3 ] P [O 3 ] = (3/5)(2/4)(1/3) (1/3) = 3/10. (d) What is the probability that the last tile is odd given that the first two are? It s really the same probability whether we are interested in the case that the first two and the last one are odd or the case that the first three are odd. Thus the probability is 3/10
6 ECE 302 Assignment 2 Solutions January 29, Repeat the previous problem with only four tiles. (a) What is P [E 3 E 1 ]? We ll need the probability that both are even. There are 2 possible sequences in which the first and third are both even. There are 4! possible sequences, thus P [E 3 E 1 ] = 2/24 = 1/12. Now P [E 3 E 1 ] = P [E 3E 1 ] P [E 1 ] = 1/12 2/4 = 1 6. (b) What is the probability that the second tile drawn is odd given that the first is? Let O j be the event that the jth tile drawn is odd. Then P [ ] = P [O 2O 1 ] P [O 1 ] = (2/4)(1/3) 2/4 = 1 3. (c) What is the conditional probability that the first two tiles drawn are odd given that third one was? Zero. There are only two odd tiles. (d) What is the probability that the last tile is odd given that the first two are? Zero. There are only two odd tiles. 10. Cellular telephone calls are handed off from one cell (having one antenna tower) to another as the user s location and the call volumes in the surrounding cells. Characterize calls as either long (L) or brief (B) and as requiring either zero (H 0 ), one (H 1 ), or more than one (H 2 ) handoff. Suppose that the probabilities are characterized by the following table: H 0 H 1 H 2 L B (a) What is the probability that the call is brief or has at least two handoffs? P [B H 2 ] = P [B H 2] P [H 2 ] = = 2 5 (b) What is the probability that the call is brief given that there were no handoffs? P [B H 0 ] = P [B H 0] P [H 0 ] = = 1 2
7 ECE 302 Assignment 2 Solutions January 29, (c) What is the probability that the call was long given that there was one handoff? P [L H 1 ] = P [L H 1] P [H 1 ] = = 2 3 (d) What is the probability that the call had at least one handoff given that it was long? P [(H 1 H 2 ) L] = P [H 1 L]+P [H 2 L] = P [H 1 L] + P [H 2 L] P [L] P [L] where we have used the fact that H 1 and H 2 are disjoint. = = 5 6,
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