Ch. 6: Discrete Probabilityy

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

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

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.

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

P(A or B) If A and B are mutually exclusive (can’t happen together, as in the king/queen example), then P(A or B)=P(A) + P(B) If A and B are NOT mutually exclusive (can happen together, as in the king/heart example), P(A or B)=P(A) + P(B) –P(A and B)

P (A and B) For independent events: P(A and B) P(A and B) = P(A) * P(B) In General: P(A and B) = P(A) * P(B/given A)

2 cards (independent) -questions Example: Pick two cards, WITH replacement from a deck of cards, P(king and king)= P(2 hearts) =

P(A and B) Example-- Independent For independent events: P(A and B) P(A and B) = P(A) * P(B) Example: Pick two cards, WITH replacement from a deck of cards, P(king and king)= 4/52 * 4/52 = 16/2704 =.0059 P(2 hearts) = 13/52 * 13/52 =.0625

P(A and B) – Dependent (without replacement) In General: P(A and B) = P(A) * P(B/given A) Example: Pick two cards, WITHOUT replacement from a deck of cards, P(king and king)= 4/52 * 3/51 = 12/2652=.0045 P(heart and heart)= 13/52 * 12/51 = 156/2652 =.059 P(king and queen) = 4/52 * 4/51 = 16/2652

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