MATH104 Ch. 11: Probability Theory part 3

Probability Assignment Assignment by intuition – based on intuition, experience, or judgment. Assignment by relative frequency – P(A) = Relative Frequency = Assignment for equally likely outcomes

One Die Experimental Probability (Relative Frequency) – If the class rolled one die 300 times and it came up a “4” 50 times, we’d say P(4)= 50/300 – The Law of Large numbers would say that our experimental results would approximate our theoretical answer. Theoretical Probability – Sample Space (outcomes): 1, 2, 3, 4, 5, 6 – P(4) = 1/6 – P(even) = 3/6

Two Dice Experimental Probability – “Team A” problem on the experiment: If we rolled a sum of “6, 7, 8, or 9” 122 times out of 218 attempts, P(6,7,8, or 9)= 122/218= 56% – Questions: What sums are possible? – Were all sums equally likely? – Which sums were most likely and why? – Use this to develop a theoretical probability – List some ways you could get a sum of 6…

Outcomes For example, to get a sum of 6, you could get: 5, 14,23,3…

Two Dice – Theoretical Probability Each die has 6 sides. How many outcomes are there for 2 sides? (Example: “1, 1”) Should we count “4,2” and “2,4” separately?

Sample Space for 2 Dice 1, 11, 21, 31, 41,51,6 2,12,22,32,42,52,6 3,13,23,33,43,53,6 4,14,24,34,44,54,6 5,15,25,35,45,55,6 6,16,26,36,46,56,6 If Team A= 6, 7, 8, 9, find P(Team A)

Two Dice- Team A/B P(Team A)= 20/36 P(Team B) = 1 – 20/36 = 16/36 Notice that P(Team A)+P(Team B) = 1

Summary of 2 Dice (Team A/B) problem TOSSING 1 DIEP(1)P(3)P(5) P(even number) Sample Empirical Results Use P(E)= n(E) / total tosses 5/30 =.176/30 =.24/30 =.1315/30 =.5 Your Empirical Results Class Empirical Results 51/300 =.17 48/ 300 =.16 53/ 300 = /300 =.49 Theoretical Results Use P(E)= n(E) / n(S) where S=SampleSpace 1/6 =.167 3/6 = 1/2

Some Probability Rules and Facts 0<= P(A) <= 1 Think of some examples where – P(A)=0P(A) = 1 The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1

One Coin Experimental – If you tossed one coin 1000 times, and 505 times came up heads, you’d say P(H)= 505/1000 – The Law of Large Numbers would say that this fraction would approach the theoretical answer as n got larger. Theoretical – Since there are only 2 equally likely outcomes, P(H)= 1/2

Two Coins Experimental Results – P(0 heads) = – P(1 head, 1 tail)= – P(2 heads)= – Note: These all sum to 1. Questions: – Why is “1 head” more likely than “2 heads”?

Two Coins- Theoretical Answer Outcomes: TT, TH, HT, HH 12 HHH H THT THTH TTT

2 Coins- Theoretical answer P(0 heads) = 1/4 P(1 head, 1 tail)= 2/4 = 1/2 P(2 heads)= ¼ Note: sum of these outcomes is 1

Summary of 2 coin problem TOSSING 2 COINSP(2 heads, 0 tails)P(1 head, 1 tail)P(0 heads, 2 tails) Guess: how often would you expect to get each possibility? Sample Empirical Results Use P(E)= n(E) / total tosses 8/30 =.2714/30 =.478/30 =.27 Your Empirical Results Class Empirical Results 73/ 300 = / 300 =.5 76/ 300 =.253 Theoretical Results Use P(E)= n(E) / n(S) 1 /42/4 = 1/21 /4

Three Coins Are “1 head”, “2 heads”, and “3 heads” all equally likely? Which are most likely and why?

Three Coins 123 HHHHH HTHHT THHTH THTT THHTHH TTHT THTTH 2*2*2=8 outcomesTTTT

3 coins P(0 heads)= P(1 head)= P(2 heads)= P(3 heads)=

Theoretical Probabilities for 3 Coins P(0 heads)= 1/8 P(1 head)= 3/8 P(2 heads)= 3/8 P(3 heads)= 1/8 Notice: Sum is 1.

Summary of 3 coin problem TOSSING 3 COINS P(3 heads)P(2 heads)P(1 head)P(0 heads) Sample Empirical Results Use P(E)= n(E) / total tosses 3/20=.157/20 =.356/20 =.34/20 =.2 Your Empirical Results Class Empirical Results 26/ 200 = / 200 = / 200 =.38 24/ 200 =.12 Theoretical Results Use P(E)= n(E) / n(S) …

Cards 4 suits, 13 denominations; 4*13=52 cards picture = J, Q, K A JQK Heart (red) Diamond (red) Clubs (black) Spades (black)

When picking one card, find… P(heart)= P(king)= P(picture card)= P(king or queen)= P(king or heart)=

Theoretical Probabilities- Cards P(heart)= 13/52 = ¼ = 0.25 P(king)= 4/52= 1/13 P(picture card)= 12/52 = 3/13 P(king or queen)= 4/ /52 = 8/52 P(king or heart)= 4/ /52 – 1/52 = 16/52

11.6 Not, Mutually Exclusive, Odds P(E)= 1-P(E ‘ ) where E’ = not E=complement of E 1. If there is a 20% chance of snow tomorrow, what is the chance it will not snow tomorrow? 2. When choosing one card from a deck, find the probability of selecting: a. A heart b. A card that is not a heart c. A king d. A card that is not a king

P(A or B) 1. When selecting one card, find the probability of: a. king or queen b. king or a heart c. king or a 5 d. 5 or a diamond e. Picture card or a 7 f. Picture card or a red card

P(A or B) Mutually exclusive events—cannot occur together If A and B are mutually exclusive, P(A or B) = P(A) + P(B) If A and B are not mutually exclusive, P(A or B) = P(A) + P(B) – P(A and B)

Odds Basic idea: If, when drawing one card from a deck, the probability of getting a heart is ¼, then The odds in favor of drawing a heart are 1:3 and the odds against a heart are 3:1. Another example: If, when drawing one card from a deck, the probability of getting a king is 1/13, then The odds in favor of drawing a king are 1:12, and the odds against a king are 12:1. Odds to Probability if odds in favor of E are a:b, then P(E)=

Given probabilities, find odds 1. Recall probabilities a. P(heart) b. P(not a heart) c. P(king) d. P(picture card) e. P(red card) 1. Find the odds in favor of: a. A heart b. A card that is not a heart c. A king d. A picture card e. A red card

…odds 2. If the odds in favor of winning the lottery are 1:1,000,000, find the probability of winning the lottery 3. If the odds in favor of getting a certain job are 3:4, find the probability of getting the job.

11.7: And, Conditional Independent events- two events are independent events if the occurrence of either of them has no effect on the probability of the other If A and B are independent events, then P(A and B) = P(A)*P(B)

2 kids 1. Assuming it’s equally likely that boys and girls are born, in a family with 2 kids, find the probability of getting: a. 2 girls b. 2 boys

2. In a family with 3 kids, find the probability of getting: Assuming P(B)=P(G) a. 3 girls b. 3 boys c. At least 1 boy

3. In a family of 4 kids, find the probability of getting: a. 4 girls b. 4 boys c. At least 1 boy

4. Two cards 4. If you pick two cards out of a deck of cards and replace them in between picks, find: a. P( 2 red cards) b. P(2 hearts) c. P(2 kings)

Dependent events— the occurrence of one of them has an effect on the occurrence of the other If A and B are dependent, P(A and B) = P(A)*P(B, given A)

Without replacement: 1. If you pick two cards out a deck without replacement, find the probability of getting: a. 2 red cards b. 2 kings

2. pick 3 cards without replacement find the probability of getting: a. 3 red cards b. 3 kings c. A king, then a queen, then a jack (in that order)

Conditional Probability Wore seat belt No seat beltTotal Driver survived 412,368162,527574,895 Driver died Total412,878164,128577,006 Find: P(driver died)= P(driver died/given no seat belt)= P(no seat belt)= P(no seat belt/given driver died)=

Wore seat belt No seat belt Total Driver survived 412,368162,527574,895 Driver died Total412,878164,128577,006 P(driver died)= 2111/577,006 = P(driver died/given no seat belt)= 1601/164,128 =.0097 P(no seat belt)= 164,128/577,006=.028 P(no seat belt/given driver died)= 1602/2111=.76

Birthday problem What is the probability that two people in this class would have the same birth date?

Hint Let E=at least two people have the same bday What is E’ (not E) Find P(E’)=