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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
n
f
Number of Outcomes Favorable to Event ( )
Total Number of Outcomes
AP A
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, 1 4,2 3,3 …
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, 1 1, 2 1, 3 1, 4 1,5 1,62,1 2,2 2,3 2,4 2,5 2,63,1 3,2 3,3 3,4 3,5 3,64,1 4,2 4,3 4,4 4,5 4,65,1 5,2 5,3 5,4 5,5 5,66,1 6,2 6,3 6,4 6,5 6,6
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)=0 P(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
1 2H HH
HT HT
T H THT TT
2 Coins- Theoretical answer
P(0 heads) = 1/4P(1 head, 1 tail)= 2/4 = 1/2P(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 Coins1 2 3
H H HHH H T HHT
T H HTHT HTT
T H H THHT THT
T H TTH2*2*2=8 outcomes T TTT
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 2 3 4 5 6 7 8 9 10 J Q KHeart (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 + 4 /52 = 8/52• P(king or heart)= 4/52 + 13/52 – 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 belt Total
Driver survived
412,368 162,527 574,895
Driver died 510 1601 2111
Total 412,878 164,128 577,006
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
412,368 162,527 574,895
Driver died
510 1601 2111
Total 412,878 164,128 577,006
• P(driver died)= 2111/577,006 = .00366• 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