1. (Tree Diagrams) Wayne and Kai keep playing chess until one of them wins two games in a row or one of them wins three games (not necessarily in a row). (a) In what percentage of all possible cases does the game end because Wayne wins three games without winning two in a row? (b) Supposing Wayne and Kai are equally skilled at chess, so that the probability of one of them winning a particular game is 0.5. What is the probability that the game ends because Wayne wins three games without winning two in a row? (Note that the answer to the two questions is not the same: (a) asks for the percentage of all outcomes, as if all outcomes are all equally likely; but in fact in (b) they are NOT equally likely, so you'll need to attach probabilities to the branches. Do the best you can to show this tree, it doesn't have to be pretty, just clear; and then give the answer to the questions.) 2. (Tree Diagrams) Kai has $4. He plays a game with Wayne in which he bets $1 on the ip of a fair coin: if the coin lands heads, Wayne gives him $1 and if tails, he gives Wayne $1. He decides to play this game four times. (a) Draw a tree diagram showing all the possible outcomes after four games, showing the amount of money Kai has after each game (no need to make this pretty, just make it clear, in whatever format you are submitting the homework). (b) What is the probability that Kai breaks even (has $4 at the end of four games)? (c) What is the probability that Kai wins money (has more than $4 at the end)? The next three problems have to do with the random experiment of rolling two fair dice and counting the number of dots that show on both. 3. (Conditional Probability) What is the probability of a sum of 5 if (a) the second roll is not 3? (b) they land on dierent numbers? 4. (Conditional Probability) Suppose that the total of the dots showing is found to be divisible by 5. What is the probability that both of them have shown 5 dots? 5. (Independence) (a) Let A = "the sum of the dots showing on the two rolls is odd" and B = "both tosses were greater than 3." Are A and B independent? Be precise. (b) Let A = "the two rolls showed the same number" and B = "the second toss was greater than 4." Are A and B independent? Be precise. 6. (Independence) We assumed in class that the relation of Independence is symmetric, that is, A and B are independent i (if and only if) B and A are independent. Justify this assumption by proving that P(A|B) = P(A) i P(B|A) = P(B). 7. (Independence) (a) Prove that P(A) = P(AB) + P(AB'). (B' is the complement of B) (b) Prove that if A and B are independent, then so are A and B'. (c) Prove that if A and B are independent, then so are A' and B'. (Hint: use (a) to prove (b); (c) is a simply corollary of (b).) 8. (Bayes Rule) A stack of cards consists of 6 red and 5 black cards. A second stack of cards consists of 9 red cards. A stack is selected at random (i.e., 0.5 probability for each) and a card is drawn and found to be red. What is the probability that it was drawn from the rst stack? You must explicitly use Bayes' Rule to solve this. 9. (Bayes Rule) A faulty communication line for digital signals changes 1/4 of the 0's to 1's and 1/3 of the 1's to 0's. If 40% of a particular le being transmitted consists of 0's and 60% consists of 1's, what is the probability that when a 0 is received it was transmitted correctly (i.e., a 0 was transmitted)? You must rst draw a diagram and then explicitly use Bayes Rule to arrive at the solution. 10. (Based on a true story, more or less....) Wayne falls asleep watching an old Clint Eastwood movie at 4am after preparing his lecture for CS 237, and has a nightmare that goes like this: He is lying on the sidewalk after robbing a bank, in pain and mulling over how to quantify the uncertainty of his survival, when Dirty Harry walks over. Dirty Harry pulls out his 44 Magnum and puts two bullets opposite each other in the six slots in the cylinder (e.g., if you number them 1 .. 6 clockwise, he puts them in 1 and 4), spins the cylinder randomly, and, saying "The question is, are you feeling lucky, probabalistically speaking, computer science punk?" points it at Wayne's head and pulls the trigger.... "CLICK!" goes the gun (no bullet) and Dirty Harry smiles... "How about that .... Let's see if this gun is memoryless!" Without spinning the cylinder again, he points the gun at Wayne's head and pulls the trigger again. (a) What is the probability that (at least in my dream) you will have to have another instructor nish out CS 237? (b) Now, suppose that when Dirty Harry put the bullets in the gun, he put them right next to each other (e.g., in slots 1 and 2). What is the probability in this case that you will have another instructor nish teaching CS 237? (c) Suppose Dirty Harry puts the bullets in two random positions in the cylinder and we don't have any idea where they are. Now what is the probability that I will not be able to nish teaching CS 237? Hint: This has nothing to do with the memory-less property (Dirty Harry never took CS 237) .... and we could solve it by just considering what the probability is for each of the patterns (no bullet, no bullet) and (no bullet, bullet) as we go around the circle of slots in the cylinder. Here is what the cylinder would look like in case (a) before Dirty Harry randomizes the positions by spinning it: Here is what it would look like in case (b) before being randomized: In case (c), of course, the bullets can be in any two slots. You may solve this using conditional probabilities, or by any other method you wish. 1 Question 1: The diagram below K = Kai and W = Wayne a. Percentage of all possible cases does the game end because Wayne wins three games without winning two in a row. The scenarios in the diagram indicating such as situation are three: 4, 8, and 9. = 3/ 10 = 0.3 b. What is the probability that the game ends because Wayne wins three games without winning two in a row? Given each has 0.5 probability of winning. The diagram below indicatesthe probability is given by winning scenario 4, 8,9=0.5 4 +2 ( 0.55 )= Question 2: a. Below is the diagram of all possible outcomes and the amout Kai has in each game. 2 b. What is the probability that Kai breaks even (has $4 at the end of four games $ 4 occurs six 16 possible outcomes at the end of the four games. The probabilit y of breaking even at the end= 6 =0 .375 16 c. What is the probability that Kai wins money (has more than $4 at the end) 5 P ( Kai wins more than $ $ at theend ) = =0.3125 16 Question 3-5: Rolling two dice sample space 3 1 1 1,1 2 3 1, 1, 2 3 2 2,1 2, 2, 2 3 3 3,1 3, 3, 2 3 4 4,1 4, 4, , 2 3 5 5,1 5, 5, 2 3 6 6,1 6, 6, 2 3 Question 3: What is the probability of a sum of 5 if 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 a. the second roll is not 3 3 30 3 = =0.1 36 36 30 b. they land on different numbers P ( 5land different numbers ) 4 30 4 = = =0.133 36 36 30 p ( land different numbers ) Question 4: Suppose that the total of the dots showing is found to be divisible by 5. What is the probability that both of them have shown 5 dots P ( divisible 5both shown5 ) 1 3 1 = = =0.333 36 36 3 p ( divisible 5 ) Question 5: Independence a. Let A = "the sum of the dots showing on the two rolls is odd" and B = "both tosses were greater than 3." Are A and B independent? Be precise Rule for independence=P ( A B )=P ( A )P ( B ) this case , P ( A B )= 4 1 18 9 1 , P ( A )= , P ( B )= hence P ( A )P ( B )= 36 9 36 36 8 4 Since P ( A B ) P ( A )P ( B ) we conclude, the two events are not independent. b. Let A = "the two rolls showed the same number" and B = "the second toss was greater than 4." Are A and B independent? Be precise 2 1 6 12 1 this case , P ( A B )= , P ( A )= , P ( B )= hence P ( A )P ( B ) = 36 18 36 36 18 The two events are independent because P ( A B ) =P ( A )P ( B ) Question 6: We assumed in class that the relation of Independence is symmetric, that is, A and B are independent iff (if and only if) B and A are independent. Justify this assumption by proving that P(A|B) = P(A) iff P(B|A) = P(B). P ( A|B )=P ( A ) iff P ( B| A )=P ( B ) starting with : P ( A|B )=P ( A ) but P ( A|B ) = therefore P( A B) =P ( A ) P (B ) which imp lies P ( B )= but P ( A B) P(A) P( A B) =P ( B| A ) P ( A) hence P ( B )=P ( B| A ) Question 7: Independence P ( A B ) P( B) 5 a. Prove that P(A) = P(AB) + P(AB'). (B' is the complement of B) P ( AB ) + P ( A B' ) =P ( A )P ( B )+ P ( A )P ( B' ) =P ( A ) { P ( B ) + P ( B ' ) } but P ( B )+ P ( B' )=1 therefore P ( AB )+ P ( A B' )=P ( A ) (1 )=P( A) b. Prove that if A and B are independent, then so are A and B'. a above P ( A )=P ( AB )+ P ( AB ' ) by definition P ( AB )=P ( A )P ( B ) this implies P ( AB )=P ( A )( 1P ( B' ) ) =P ( A )P ( A )P ( B' ) so P ( A ) P ( B ' ) =P ( A )P ( AB ) but part a above P ( A )P ( AB )=P ( A B' ) that is P ( A ) P ( B' )=P( A B' ) c. Prove that if A and B are independent, then so are A' and B'. As corollary from part (b) : if two events independent then each event is independent of other event complement, which implies that the complements of both event are independent. Therefore, A' and B' are independent events. Question 8: A stack of cards consists of 6 red and 5 black cards. A second stack of cards consists of 9 red cards. A stack is selected at random (i.e., 0.5 probability for each) and a card is drawn and found to be red. What is the probability that it was drawn from the first stack? You must explicitly use Bayes' Rule to solve this. let A=event B=event alsolet R=event card Bayes theorem: P ( A|R )= P ( R|A ) P( A) P ( R) 6 P ( A )=0.5, P ( R )=P ( R| A ) P ( A ) + P ( R|B ) P ( B )= 6 0.5+1 0.5=0.773 11 6 11 therefore : P ( A|R ) = 0.5=0.353 0.773 Question 9: A faulty communication line for digital signals changes 1/4 of the 0's to 1's and 1/3 of the 1's to 0's. If 40% of a particular file being transmitted consists of 0's and 60% consists of 1's, what is the probability that when a 0 is received it was transmitted correctly (i.e., a 0 was transmitted)? You must first draw a diagram and then explicitly use Bayes Rule to arrive at the solution. Let A=event 0B=1, W =event 0 transmitted P ( A ) =0.4P ( B )=0.6 1 3 1 P (W |A )=1 = , P ( W |B )= , P ( W )=P ( W | A ) P ( A ) + P ( W |B ) P ( B ) 4 4 3 P (W ) = 0.43 0.61 + =0.5 4 3 P ( A|W )= P ( W |A ) 0.75 P ( A )= 0.4=0.6 0.5 P (W ) Question 10: a. What is the probability that (at least in my dream) you will have to have another instructor finish out CS 237? 2 1 = 6 3 7 b. Now, suppose that when Dirty Harry put the bullets in the gun, he put them right next to each other (e.g., in slots 1 and 2). What is the probability in this case that you will have another instructor finish teaching CS 237? 2 1 = 6 3 c. Suppose Dirty Harry puts the bullets in two random positions in the cylinder and we don't have any idea where they are. Now what is the probability that I will not be able to finish teaching CS 237? 2 1 = 6 3 1 Question 1: The diagram below K = Kai and W = Wayne a. Percentage of all possible cases does the game end because Wayne wins three games without winning two in a row. The scenarios in the diagram indicating such as situation are three: 4, 8, and 9. = 3/ 10 = 0.3 b. What is the probability that the game ends because Wayne wins three games without winning two in a row? Given each has 0.5 probability of winning. The diagram below indicatesthe probability is given by winning scenario 4, 8,9=0.5 4 +2 ( 0.55 )= Question 2: a. Below is the diagram of all possible outcomes and the amout Kai has in each game. 2 b. What is the probability that Kai breaks even (has $4 at the end of four games $ 4 occurs six 16 possible outcomes at the end of the four games. The probabilit y of breaking even at the end= 6 =0 .375 16 c. What is the probability that Kai wins money (has more than $4 at the end) 5 P ( Kai wins more than $ $ at theend ) = =0.3125 16 Question 3-5: Rolling two dice sample space 3 1 1 1,1 2 3 1, 1, 2 3 2 2,1 2, 2, 2 3 3 3,1 3, 3, 2 3 4 4,1 4, 4, , 2 3 5 5,1 5, 5, 2 3 6 6,1 6, 6, 2 3 Question 3: What is the probability of a sum of 5 if 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 a. the second roll is not 3 3 30 3 = =0.1 36 36 30 b. they land on different numbers P ( 5land different numbers ) 4 30 4 = = =0.133 36 36 30 p ( land different numbers ) Question 4: Suppose that the total of the dots showing is found to be divisible by 5. What is the probability that both of them have shown 5 dots P ( divisible 5both shown5 ) 1 3 1 = = =0.333 36 36 3 p ( divisible 5 ) Question 5: Independence a. Let A = "the sum of the dots showing on the two rolls is odd" and B = "both tosses were greater than 3." Are A and B independent? Be precise Rule for independence=P ( A B )=P ( A )P ( B ) this case , P ( A B )= 4 1 18 9 1 , P ( A )= , P ( B )= hence P ( A )P ( B )= 36 9 36 36 8 4 Since P ( A B ) P ( A )P ( B ) we conclude, the two events are not independent. b. Let A = "the two rolls showed the same number" and B = "the second toss was greater than 4." Are A and B independent? Be precise 2 1 6 12 1 this case , P ( A B )= , P ( A )= , P ( B )= hence P ( A )P ( B ) = 36 18 36 36 18 The two events are independent because P ( A B ) =P ( A )P ( B ) Question 6: We assumed in class that the relation of Independence is symmetric, that is, A and B are independent iff (if and only if) B and A are independent. Justify this assumption by proving that P(A|B) = P(A) iff P(B|A) = P(B). P ( A|B )=P ( A ) iff P ( B| A )=P ( B ) starting with : P ( A|B )=P ( A ) but P ( A|B ) = therefore P( A B) =P ( A ) P (B ) which imp lies P ( B )= but P ( A B) P(A) P( A B) =P ( B| A ) P ( A) hence P ( B )=P ( B| A ) Question 7: Independence P ( A B ) P( B) 5 a. Prove that P(A) = P(AB) + P(AB'). (B' is the complement of B) P ( AB ) + P ( A B' ) =P ( A )P ( B )+ P ( A )P ( B' ) =P ( A ) { P ( B ) + P ( B ' ) } but P ( B )+ P ( B' )=1 therefore P ( AB )+ P ( A B' )=P ( A ) (1 )=P( A) b. Prove that if A and B are independent, then so are A and B'. a above P ( A )=P ( AB )+ P ( AB ' ) by definition P ( AB )=P ( A )P ( B ) this implies P ( AB )=P ( A )( 1P ( B' ) ) =P ( A )P ( A )P ( B' ) so P ( A ) P ( B ' ) =P ( A )P ( AB ) but part a above P ( A )P ( AB )=P ( A B' ) that is P ( A ) P ( B' )=P( A B' ) c. Prove that if A and B are independent, then so are A' and B'. As corollary from part (b) : if two events independent then each event is independent of other event complement, which implies that the complements of both event are independent. Therefore, A' and B' are independent events. Question 8: A stack of cards consists of 6 red and 5 black cards. A second stack of cards consists of 9 red cards. A stack is selected at random (i.e., 0.5 probability for each) and a card is drawn and found to be red. What is the probability that it was drawn from the first stack? You must explicitly use Bayes' Rule to solve this. let A=event B=event alsolet R=event card Bayes theorem: P ( A|R )= P ( R|A ) P( A) P ( R) 6 P ( A )=0.5, P ( R )=P ( R| A ) P ( A ) + P ( R|B ) P ( B )= 6 0.5+1 0.5=0.773 11 6 11 therefore : P ( A|R ) = 0.5=0.353 0.773 Question 9: A faulty communication line for digital signals changes 1/4 of the 0's to 1's and 1/3 of the 1's to 0's. If 40% of a particular file being transmitted consists of 0's and 60% consists of 1's, what is the probability that when a 0 is received it was transmitted correctly (i.e., a 0 was transmitted)? You must first draw a diagram and then explicitly use Bayes Rule to arrive at the solution. Let A=event 0B=1, W =event 0 transmitted P ( A ) =0.4P ( B )=0.6 1 3 1 P (W |A )=1 = , P ( W |B )= , P ( W )=P ( W | A ) P ( A ) + P ( W |B ) P ( B ) 4 4 3 P (W ) = 0.43 0.61 + =0.5 4 3 P ( A|W )= P ( W |A ) 0.75 P ( A )= 0.4=0.6 0.5 P (W ) Question 10: a. What is the probability that (at least in my dream) you will have to have another instructor finish out CS 237? 2 1 = 6 3 7 b. Now, suppose that when Dirty Harry put the bullets in the gun, he put them right next to each other (e.g., in slots 1 and 2). What is the probability in this case that you will have another instructor finish teaching CS 237? 2 1 = 6 3 c. Suppose Dirty Harry puts the bullets in two random positions in the cylinder and we don't have any idea where they are. Now what is the probability that I will not be able to finish teaching CS 237? 2 1 = 6 3 1 Question 1: The diagram below K = Kai and W = Wayne a. Percentage of all possible cases does the game end because Wayne wins three games without winning two in a row. The scenarios in the diagram indicating such as situation are three: 4, 8, and 9. = 3/ 10 = 0.3 b. What is the probability that the game ends because Wayne wins three games without winning two in a row? Given each has 0.5 probability of winning. The diagram below indicatesthe probability is given by winning scenario 4, 8,9=0.5 4 +2 ( 0.55 )= Question 2: a. Below is the diagram of all possible outcomes and the amout Kai has in each game. 2 b. What is the probability that Kai breaks even (has $4 at the end of four games $ 4 occurs six 16 possible outcomes at the end of the four games. The probabilit y of breaking even at the end= 6 =0 .375 16 c. What is the probability that Kai wins money (has more than $4 at the end) 5 P ( Kai wins more than $ $ at theend ) = =0.3125 16 Question 3-5: Rolling two dice sample space 3 1 1 1,1 2 3 1, 1, 2 3 2 2,1 2, 2, 2 3 3 3,1 3, 3, 2 3 4 4,1 4, 4, , 2 3 5 5,1 5, 5, 2 3 6 6,1 6, 6, 2 3 Question 3: What is the probability of a sum of 5 if 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 a. the second roll is not 3 3 30 3 = =0.1 36 36 30 b. they land on different numbers P ( 5land different numbers ) 4 30 4 = = =0.133 36 36 30 p ( land different numbers ) Question 4: Suppose that the total of the dots showing is found to be divisible by 5. What is the probability that both of them have shown 5 dots P ( divisible 5both shown5 ) 1 3 1 = = =0.333 36 36 3 p ( divisible 5 ) Question 5: Independence a. Let A = "the sum of the dots showing on the two rolls is odd" and B = "both tosses were greater than 3." Are A and B independent? Be precise Rule for independence=P ( A B )=P ( A )P ( B ) this case , P ( A B )= 4 1 18 9 1 , P ( A )= , P ( B )= hence P ( A )P ( B )= 36 9 36 36 8 4 Since P ( A B ) P ( A )P ( B ) we conclude, the two events are not independent. b. Let A = "the two rolls showed the same number" and B = "the second toss was greater than 4." Are A and B independent? Be precise 2 1 6 12 1 this case , P ( A B )= , P ( A )= , P ( B )= hence P ( A )P ( B ) = 36 18 36 36 18 The two events are independent because P ( A B ) =P ( A )P ( B ) Question 6: We assumed in class that the relation of Independence is symmetric, that is, A and B are independent iff (if and only if) B and A are independent. Justify this assumption by proving that P(A|B) = P(A) iff P(B|A) = P(B). P ( A|B )=P ( A ) iff P ( B| A )=P ( B ) starting with : P ( A|B )=P ( A ) but P ( A|B ) = therefore P( A B) =P ( A ) P (B ) which imp lies P ( B )= but P ( A B) P(A) P( A B) =P ( B| A ) P ( A) hence P ( B )=P ( B| A ) Question 7: Independence P ( A B ) P( B) 5 a. Prove that P(A) = P(AB) + P(AB'). (B' is the complement of B) P ( AB ) + P ( A B' ) =P ( A )P ( B )+ P ( A )P ( B' ) =P ( A ) { P ( B ) + P ( B ' ) } but P ( B )+ P ( B' )=1 therefore P ( AB )+ P ( A B' )=P ( A ) (1 )=P( A) b. Prove that if A and B are independent, then so are A and B'. a above P ( A )=P ( AB )+ P ( AB ' ) by definition P ( AB )=P ( A )P ( B ) this implies P ( AB )=P ( A )( 1P ( B' ) ) =P ( A )P ( A )P ( B' ) so P ( A ) P ( B ' ) =P ( A )P ( AB ) but part a above P ( A )P ( AB )=P ( A B' ) that is P ( A ) P ( B' )=P( A B' ) c. Prove that if A and B are independent, then so are A' and B'. As corollary from part (b) : if two events independent then each event is independent of other event complement, which implies that the complements of both event are independent. Therefore, A' and B' are independent events. Question 8: A stack of cards consists of 6 red and 5 black cards. A second stack of cards consists of 9 red cards. A stack is selected at random (i.e., 0.5 probability for each) and a card is drawn and found to be red. What is the probability that it was drawn from the first stack? You must explicitly use Bayes' Rule to solve this. let A=event B=event alsolet R=event card Bayes theorem: P ( A|R )= P ( R|A ) P( A) P ( R) 6 P ( A )=0.5, P ( R )=P ( R| A ) P ( A ) + P ( R|B ) P ( B )= 6 0.5+1 0.5=0.773 11 6 11 therefore : P ( A|R ) = 0.5=0.353 0.773 Question 9: A faulty communication line for digital signals changes 1/4 of the 0's to 1's and 1/3 of the 1's to 0's. If 40% of a particular file being transmitted consists of 0's and 60% consists of 1's, what is the probability that when a 0 is received it was transmitted correctly (i.e., a 0 was transmitted)? You must first draw a diagram and then explicitly use Bayes Rule to arrive at the solution. Let A=event 0B=1, W =event 0 transmitted P ( A ) =0.4P ( B )=0.6 1 3 1 P (W |A )=1 = , P ( W |B )= , P ( W )=P ( W | A ) P ( A ) + P ( W |B ) P ( B ) 4 4 3 P (W ) = 0.43 0.61 + =0.5 4 3 P ( A|W )= P ( W |A ) 0.75 P ( A )= 0.4=0.6 0.5 P (W ) Question 10: a. What is the probability that (at least in my dream) you will have to have another instructor finish out CS 237? 2 1 = 6 3 7 b. Now, suppose that when Dirty Harry put the bullets in the gun, he put them right next to each other (e.g., in slots 1 and 2). What is the probability in this case that you will have another instructor finish teaching CS 237? 2 1 = 6 3 c. Suppose Dirty Harry puts the bullets in two random positions in the cylinder and we don't have any idea where they are. Now what is the probability that I will not be able to finish teaching CS 237? 2 1 = 6 3 1 Question 1: The diagram below K = Kai and W = Wayne a. Percentage of all possible cases does the game end because Wayne wins three games without winning two in a row. The scenarios in the diagram indicating such as situation are three: 4, 8, and 9. = 3/ 10 = 0.3 b. What is the probability that the game ends because Wayne wins three games without winning two in a row? Given each has 0.5 probability of winning. The diagram below indicatesthe probability is given by winning scenario 4, 8,9=0.5 4 +2 ( 0.55 )= Question 2: a. Below is the diagram of all possible outcomes and the amout Kai has in each game. 2 b. What is the probability that Kai breaks even (has $4 at the end of four games $ 4 occurs six 16 possible outcomes at the end of the four games. The probabilit y of breaking even at the end= 6 =0 .375 16 c. What is the probability that Kai wins money (has more than $4 at the end) 5 P ( Kai wins more than $ $ at theend ) = =0.3125 16 Question 3-5: Rolling two dice sample space 3 1 1 1,1 2 3 1, 1, 2 3 2 2,1 2, 2, 2 3 3 3,1 3, 3, 2 3 4 4,1 4, 4, , 2 3 5 5,1 5, 5, 2 3 6 6,1 6, 6, 2 3 Question 3: What is the probability of a sum of 5 if 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 a. the second roll is not 3 3 30 3 = =0.1 36 36 30 b. they land on different numbers P ( 5land different numbers ) 4 30 4 = = =0.133 36 36 30 p ( land different numbers ) Question 4: Suppose that the total of the dots showing is found to be divisible by 5. What is the probability that both of them have shown 5 dots P ( divisible 5both shown5 ) 1 3 1 = = =0.333 36 36 3 p ( divisible 5 ) Question 5: Independence a. Let A = "the sum of the dots showing on the two rolls is odd" and B = "both tosses were greater than 3." Are A and B independent? Be precise Rule for independence=P ( A B )=P ( A )P ( B ) this case , P ( A B )= 4 1 18 9 1 , P ( A )= , P ( B )= hence P ( A )P ( B )= 36 9 36 36 8 4 Since P ( A B ) P ( A )P ( B ) we conclude, the two events are not independent. b. Let A = "the two rolls showed the same number" and B = "the second toss was greater than 4." Are A and B independent? Be precise 2 1 6 12 1 this case , P ( A B )= , P ( A )= , P ( B )= hence P ( A )P ( B ) = 36 18 36 36 18 The two events are independent because P ( A B ) =P ( A )P ( B ) Question 6: We assumed in class that the relation of Independence is symmetric, that is, A and B are independent iff (if and only if) B and A are independent. Justify this assumption by proving that P(A|B) = P(A) iff P(B|A) = P(B). P ( A|B )=P ( A ) iff P ( B| A )=P ( B ) starting with : P ( A|B )=P ( A ) but P ( A|B ) = therefore P( A B) =P ( A ) P (B ) which imp lies P ( B )= but P ( A B) P(A) P( A B) =P ( B| A ) P ( A) hence P ( B )=P ( B| A ) Question 7: Independence P ( A B ) P( B) 5 a. Prove that P(A) = P(AB) + P(AB'). (B' is the complement of B) P ( AB ) + P ( A B' ) =P ( A )P ( B )+ P ( A )P ( B' ) =P ( A ) { P ( B ) + P ( B ' ) } but P ( B )+ P ( B' )=1 therefore P ( AB )+ P ( A B' )=P ( A ) (1 )=P( A) b. Prove that if A and B are independent, then so are A and B'. a above P ( A )=P ( AB )+ P ( AB ' ) by definition P ( AB )=P ( A )P ( B ) this implies P ( AB )=P ( A )( 1P ( B' ) ) =P ( A )P ( A )P ( B' ) so P ( A ) P ( B ' ) =P ( A )P ( AB ) but part a above P ( A )P ( AB )=P ( A B' ) that is P ( A ) P ( B' )=P( A B' ) c. Prove that if A and B are independent, then so are A' and B'. As corollary from part (b) : if two events independent then each event is independent of other event complement, which implies that the complements of both event are independent. Therefore, A' and B' are independent events. Question 8: A stack of cards consists of 6 red and 5 black cards. A second stack of cards consists of 9 red cards. A stack is selected at random (i.e., 0.5 probability for each) and a card is drawn and found to be red. What is the probability that it was drawn from the first stack? You must explicitly use Bayes' Rule to solve this. let A=event B=event alsolet R=event card Bayes theorem: P ( A|R )= P ( R|A ) P( A) P ( R) 6 P ( A )=0.5, P ( R )=P ( R| A ) P ( A ) + P ( R|B ) P ( B )= 6 0.5+1 0.5=0.773 11 6 11 therefore : P ( A|R ) = 0.5=0.353 0.773 Question 9: A faulty communication line for digital signals changes 1/4 of the 0's to 1's and 1/3 of the 1's to 0's. If 40% of a particular file being transmitted consists of 0's and 60% consists of 1's, what is the probability that when a 0 is received it was transmitted correctly (i.e., a 0 was transmitted)? You must first draw a diagram and then explicitly use Bayes Rule to arrive at the solution. Let A=event 0B=1, W =event 0 transmitted P ( A ) =0.4P ( B )=0.6 1 3 1 P (W |A )=1 = , P ( W |B )= , P ( W )=P ( W | A ) P ( A ) + P ( W |B ) P ( B ) 4 4 3 P (W ) = 0.43 0.61 + =0.5 4 3 P ( A|W )= P ( W |A ) 0.75 P ( A )= 0.4=0.6 0.5 P (W ) Question 10: a. What is the probability that (at least in my dream) you will have to have another instructor finish out CS 237? 2 1 = 6 3 7 b. Now, suppose that when Dirty Harry put the bullets in the gun, he put them right next to each other (e.g., in slots 1 and 2). What is the probability in this case that you will have another instructor finish teaching CS 237? 2 1 = 6 3 c. Suppose Dirty Harry puts the bullets in two random positions in the cylinder and we don't have any idea where they are. Now what is the probability that I will not be able to finish teaching CS 237? 2 1 = 6 3