Probability: Part 1 Naima Hammoud
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1 Probability: Part 1 Naima ammoud Feb 7, 2017
2 Motivation ossing a coin Rolling a die Outcomes: eads or ails Outcomes: 1, 2, 3, 4, 5 or 6
3 Defining Probability If I toss a coin, there is a 50% chance I will get eads. If I roll a die, there is a 1 in 6 chance I will get a 3. If I look at the weather app, it says there is a 90% chance of rain.
4 Defining Probability If I toss a coin, there is a 50% chance I will get eads. If I roll a die, there is a 1 in 6 chance I will get a 3. If I look at the weather app, it says there is a 90% chance of rain. A probability is a measure of the chance that an event is going to happen
5 Properties of Probability It is a positive number between 0 and 1 A probability of 0 means that there is 0% chance that an event will happen, i.e. we are sure the event will not happen A probability of 1 means that we are 100% sure an event will happen Multiplying the probability by 100 gives a percentage he probability of an event happening is denoted by: P(event) = a number between 0 and 1
6 Definitions Sample Space: the collection of events that contains every possible outcome. Example: the sample space in a coin toss is {eads, ails} the sample space in rolling a die is {1, 2, 3, 4, 5, 6}
7 Definitions Sample Space: the collection of events that contains every possible outcome. Example: the sample space in a coin toss is {eads, ails} the sample space in rolling a die is {1, 2, 3, 4, 5, 6} Event: a specific outcome of the sample space. Example: observing eads when tossing a coin observing a 4 when rolling a die
8 Probability Examples If I flip a fair coin, then I can define the following events: : the event that eads is observed : the event that ails is observed P() = 0.5 and P() = 0.5 If I roll a die, I can define the following event: A: the event of rolling a number larger than 3 P(A) = P(rolling 4 or 5 or 6) = 3/6 = 1/2
9 More Definitions Probabilities always sum up to 1 When rolling a die, the sample space is {1, 2, 3, 4, 5, 6} If A: the event that 5 or 6 occur, then P(A)=1/3 he complement of A is denoted by A c, where A c : event that 1, 2, 3 or 4 occur P(A c ) = 2/3 because P(A)+ P(A c ) = 1
10 More Definitions Independent events Example: if I toss a coin, each toss is independent of the one that preceded it. If I get 10 heads in a row, the probability to observe a head on the 11 th toss is still 0.5 (or 50%) Disjoint events When two or more events have nothing in common. Example: rolling a six-sided die A: I roll a die and observe a 3 B: I roll a die and observe a 1 or a 6
11 More Definitions Independent events Example: if I toss a coin, each toss is independent of the one that preceded it. If I get 10 heads in a row, the probability to observe a head on the 11 th toss is still 0.5 (or 50%) Disjoint events When two or more events have nothing in common. Example: rolling a six-sided die A: I roll a die and observe a 3 A and B are disjoint B: I roll a die and observe a 1 or a 6
12 Examples 1. ossing two heads in a row 2. ossing three tails in a row 3. ossing a head, a tail, followed by another tail 4. Rolling a 5 then rolling a 6 5. Rolling a 5 or a 6 6. Rolling a 1, 2, or 3 on the first roll, then 4, 5, or 6 on the second roll
13 Examples 1. ossing two heads in a row 2. ossing three tails in a row 3. ossing a head, a tail, followed by another tail 4. Rolling a 5 then rolling a 6 5. Rolling a 5 or a 6 6. Rolling a 1, 2, or 3 on the first roll, then 4, 5, or 6 on the second roll Rule of thumb: OR + disjoint = add AND + independent = multiply
14 ossing wo Coins able representation Outcomes: {,,, }
15 ossing wo Coins able representation ree representation () () () () Outcomes: {,,, }
16 ossing wo Coins able representation ree representation () () () () Outcomes: {,,, } P ()=
17 ossing wo Coins able representation ree representation () () () () Outcomes: {,,, } P ()=
18 ossing wo Coins able representation ree representation () () () () Outcomes: {,,, } P ()=
19 ossing wo Coins able representation ree representation () () () () Outcomes: {,,, } P ()=0.25
20 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head?
21 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head?
22 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head? P (at least one head) = P ( or or )
23 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head? P (at least one head) = P ()+P ()+P ()
24 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head? P (at least one head) = P ()+P ()+P () 1/4 1/4 1/4
25 ossing wo Coins otal of 4 outcomes Each outcome has a 1 in 4 or 0.25 (or 25%) chance of happening What is the probability of obtaining at least one head? P (at least one head) = P ()+P ()+P () =3/4 Is there a difference between flipping two coins or flipping the same coin twice?
26 ossing hree Coins 1
27 ossing hree Coins 1 2
28 ossing hree Coins 1 2 3
29 ossing hree Coins Sample Space: {,,,,,,, }
30 ossing Multiple Coins 1 coin: 2 outcomes
31 ossing Multiple Coins 1 coin: 2 outcomes 2 coins: 4 outcomes
32 ossing Multiple Coins 1 coin: 2 outcomes 2 coins: 4 outcomes 3 coins: 8 outcomes
33 ossing Multiple Coins 1 coin: 2 outcomes 2 coins: 4 outcomes 3 coins: 8 outcomes 4 coins?
34 ossing Multiple Coins With each coin toss there are WO possibilities: eads and ails 1 coin: 2 outcomes 2 coins: 4 outcomes 3 coins: 8 outcomes 4 coins?
35 ossing Multiple Coins With each coin toss there are WO possibilities: eads and ails 1 coin: 2 outcomes 2 coins: 4 outcomes coins: 8 outcomes coins? 2 4 = 16 outcomes
36 Rolling wo Dice With each roll of a die there are SIX possibilities: 1, 2, 3, 4, 5, 6 1 roll: 6 outcomes 2 rolls: 6 2 = 36 outcomes A tree representation will look quite messy A better representation is with a table
37 Rolling wo Dice ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1,
38 Rolling wo Dice ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1, outcomes P(each outcome) = 1/36 P(sum = 2) =? P(sum = 4) =? P(sum = 7) =? P(obtaining at least one 2) =? Careful about double counting
39 Rolling wo Dice ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1, outcomes P(each outcome) = 1/36 P(sum = 2) =? P(sum = 4) =? P(sum = 7) =? P(obtaining at least one 2) =? Careful about double counting What is the probability that I get a 3 on the second roll, knowing that I got a 5 on the first?
40 Introducing Conditional Probability ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1, outcomes P(each outcome) = 1/36 P(sum = 2) =? P(sum = 4) =? P(sum = 7) =? P(obtaining at least one 2) =? Careful about double counting What is the probability that I get a 3 on the second roll, knowing that I got a 5 on the first?
41 Introducing Conditional Probability ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1, outcomes P(each outcome) = 1/36 P(sum = 2) =? P(sum = 4) =? P(sum = 7) =? P(obtaining at least one 2) =? Careful about double counting What is the probability that I get a 3 on the second roll, knowing that I got a 5 on the first?
42 Introducing Conditional Probability ,1 6,2 6,3 6,4 6,5 6,6 5,1 5,2 5,3 5,4 5,5 5,6 4,1 4,2 4,3 4,4 4,5 4,6 3,1 3,2 3,3 3,4 3,5 3,6 2,1 2,2 2,3 2,4 2,5 2,6 1,1 1,2 1,3 1,4 1,5 1, outcomes P(each outcome) = 1/36 P(sum = 2) =? P(sum = 4) =? P(sum = 7) =? P(obtaining at least one 2) =? Careful about double counting What is the probability that I get a 3 on the second roll, knowing that I got a 5 on the first? 1/6
43 Conditional Probability Monty all
44 Conditional Probability Formal Definition Probability of event B happening given that event A has already occurred P (event B given event A) = P (B A) P (B A) = P (A and B) P (B)
45 Conditional Probability Formal Definition Probability of event B happening given that event A has already occurred P (event B given event A) = P (B A) P (B A) = P (A and B) P (B) P (A and B) = P (B A)P (B) =P (B and A)
46 Conditional Probability Formal Definition Probability of event B happening given that event A has already occurred P (event B given event A) = P (B A) P (B A) = P (A and B) P (B) P (A and B) = P (B A)P (B) =P (B and A) P (A B)P (A)
47 Conditional Probability Bayes heorem Probability of event B happening given that event A has already occurred P (event B given event A) = P (B A) P (B A) = P (A and B) P (B) = P (A B)P (A) P (B) Bayes heorem
48 Conditional Probability Monty all Solution C1 = car behind door 1 D1 = host opens door 1 C2 = car behind door 2 D2 = host opens door 2 C3 = car behind door 3 D3 = host opens door 3 P(C1) = P(C2) = P(C3) = 1/3 I choose door 1, so now I know the host has to open either door 2 or door 3 with equal probability, but the host knows where the car is, so: P(D2 C1) = 0.5, P(D2 C2) = 0, P(D2 C3) =1 P(D3 C1) = 0.5, P(D3 C2) = 1, P(D3 C3) = 0
49 Conditional Probability Monty all Solution P(C1) = P(C2) = P(C3) = 1/3 I choose door 1 P(D2 C1) = 0.5, P(D2 C2) = 0, P(D2 C3) =1 P(D3 C1) = 0.5, P(D3 C2) = 1, P(D3 C3) = 0 ost opens D3 and it has a goat Compare P(C1 D3) and P(C2 D3)
50 Conditional Probability Monty all Solution P(C1) = P(C2) = P(C3) = 1/3 I choose door 1 P(D2 C1) = 0.5, P(D2 C2) = 0, P(D2 C3) =1 P(D3 C1) = 0.5, P(D3 C2) = 1, P(D3 C3) = 0 ost opens D3 and it has a goat Compare P(C1 D3) and P(C2 D3) P (C1 D3) = P (D3 C1)P (C1) P (D3) P (C2 D3) = P (D3 C2)P (C2) P (D3)
51 Conditional Probability Monty all Solution P(C1) = P(C2) = P(C3) = 1/3 I choose door 1 P(D2 C1) = 0.5, P(D2 C2) = 0, P(D2 C3) =1 P(D3 C1) = 0.5, P(D3 C2) = 1, P(D3 C3) = 0 ost opens D3 and it has a goat Compare P(C1 D3) and P(C2 D3) P (C1 D3) = P (D3 C1)P (C1) P (D3) P (C2 D3) = P (D3 C2)P (C2) P (D3) 1/3 2/3
52 Conditional Probability Monty all Solution SWIC O DOOR 2 P (C1 D3) = P (D3 C1)P (C1) P (D3) P (C2 D3) = 1/3 2/3 P (D3 C2)P (C2) P (D3)
53 Conditional Probability Radar Example Suppose that the probability you see a plane when you look up to the sky is 5%, and let s say that we have a radar which registers a plane with 99% accuracy. here is also a 10% chance for the radar to sound a false alarm. A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c here is a plane and radar registered it here is a plane, but radar did not register it 0.95 A c B B c False alarm here is no plane and radar did not register anything
54 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c 0.95 A c B B c What is the probability that the radar registers something?
55 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c 0.95 A c B B c What is the probability that the radar registers something? here is a plane and the radar registers it
56 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c 0.95 A c B B c What is the probability that the radar registers something? here is a plane and the radar registers it here is no plane but the radar makes a false alarm
57 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c = A c B B c = What is the probability that the radar registers something? here is a plane and the radar registers it here is no plane but the radar makes a false alarm
58 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c = A c B B c = What is the probability that the radar registers something? P (B) =P (A and B)+P (A c and B) = = here is a 14.45% chance that the radar registers something!
59 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c = A c B B c = What is the probability that there is a plane given that the radar registered something? P (A B) = P (A and B) P (B) = =0.34
60 Conditional Probability Radar Example A = event that we see a plane (A c is the event that we don t see a plane) B = event that the radar registers something (B c means radar didn't register) 0.05 A B B c = A c B B c = What is the probability that there is a plane given that the radar registered something? If the radar registers P (A and B) P (A B) = something, there s only = P (B) =0.34 a 34% chance that there s actually a plane!
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