Ch. 6: Discrete Probability-- Questions

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)= _____ – 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) = ____ – P(even) = ___

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:

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 List the outcomes in the sample space If Team A= 6, 7, 8, 9, find P(Team A)

Two Dice- Team A/B P(Team A)= ___ P(Team B) = ___ Notice that P(Team A)+P(Team B) = ___

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)= ___ – 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)= ___

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:

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

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

Three Coins

3 coins P(0 heads)= P(1 head)= P(2 heads)= P(3 heads)= Note: sum is ____

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)=

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)= ___ P(2 hearts) = ____

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)= ____ P(heart and heart)= ____ P(king and queen) = ___

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)= ___ P(driver died/given no seat belt)= ___ P(no seat belt)= ___ P(no seat belt/given driver died)= ___