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