Math 1710 Class 4pi.math.cornell.edu/~web1710/slides/aug31_v1.pdf · Math 1710 Class 4 V1 Formal...

Math 1710Class 4

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Smallest ofTwo a Queen

ConditionalProbability

Independence

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Disjoint vs.Independent

BeginningTree Diagrams

Math 1710 Class 4

Dr. Allen Back

Aug. 31, 2016

Math 1710Class 4

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Fancy Words

Random Phenomenon - Individually not predictable.But if repeated, long term relative frequency is.

Math 1710Class 4

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Fancy Words

Random PhenomenonOutcome

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. rolling a die

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. rolling a dieOutcomee.g. a 3.

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. rolling a dieOutcomee.g. a 3.Sample Space S - Set of all possible outcomes.e.g. S = {1, 2, 3, 4, 5, 6}.

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. rolling a dieOutcomee.g. a 3.Sample Space S - Set of all possible outcomes.e.g. S = {1, 2, 3, 4, 5, 6}.Event E - a subset of the sample space – a set of outcomes.e.g. E = {2, 4, 6}“Rolling an even number.”

Math 1710Class 4

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Fancy Words

Random Phenomenon - Individually not predictable.But if repeated, long term relative frequency is.Random Phenomenon -e.g. tossing two coins

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. tossing two coinsOutcomee.g. HT.

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. tossing two coinsOutcomee.g. HT.Sample Space S - Set of all possible outcomes.e.g. S = {HH,HT ,TH,TT}.

Math 1710Class 4

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Fancy Words

Random Phenomenon -e.g. tossing two coinsOutcomee.g. HT.Sample Space S - Set of all possible outcomes.e.g. S = {HH,HT ,TH,TT}.Event E - a subset of the sample space – a set of outcomes.e.g. E = {HT ,TH}“Rolling a total of 1 head.”

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Probability Assignment on a Sample Space S

We won’t usually be this formal!(1) 0 ≤ P(Outcome) ≤ 1.(2) Σ P(All Outcomes in S)= 1.(3) Σ P(Outcomes in an Event)=P(Event).

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Probability Properties

(1) 0 ≤ P(E ) ≤ 1 for any event E .(2) P(φ) = 0, P(S) = 1.(3) P(A ∪ B) = P(A) + P(B) if A and B are disjoint.(4) P(Ac) = 1− P(A)

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A Probability Assignment

English language frequencies:E 13.0% H 3.5% W 1.6T 9.3% L 3.5% V 1.3%N 7.8% C 3.0% B 0.9%R 7.7% F 2.8% X 0.5%O 7.4% P 2.7% K 0.3%I 7.4% U 2.7% Q 0.3%A 7.3% M 2.5% J 0.2%S 6.3% Y 1.9% Z 0.1%D 4.4% G 1.6%

Sample Space S: S={A,B,C,...Z}.An Outcome: e.g. an M.An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.

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Sample Space S: S={A,B,C,...Z}.

An Outcome: e.g. an M.An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.

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Sample Space S: S={A,B,C,...Z}.An Outcome: e.g. an M.

An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.

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Math 1710Class 4

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Prob of Event frm Prob Outcomes


E2: a letter in ‘HELLO”.E2 = {H,E , L,O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.

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Sample Space S: S={A,B,C,...Z}.An Outcome: e.g. an M.An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.E2: a letter in ‘HELLO”.

E2 = {H,E , L,O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.

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Sample Space S: S={A,B,C,...Z}.An Outcome: e.g. an M.An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.E2: a letter in ‘HELLO”.E2 = {H,E , L,O}.

P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.

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Sample Space S: S={A,B,C,...Z}.An Outcome: e.g. an M.An Event E: e.g. being a vowel. Formally: E={A,E,I,O,U}P(E ) = 13.0 + 7.4 + 7.4 + 7.3 + 2.7% = 37.8%.E2: a letter in ‘HELLO”.E2 = {H,E , L,O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.

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Addition Rule

E2 = {H,E , L.O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.E3 = {B,A,D}.P(E3) = 0.9 + 7.3 + 4.4% = 12.6%.E2 and E3 are disjoint events.

P(E2 ∪ E3) = P(E2) + P(E3) = 27.4 + 12.6%.E2 ∪ E3 means “a letter in HELLO or BAD.”E2 ∪ E3 = {H,E , L,O,B,A,D}.Union (∪) corresponds to “or’.’

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Addition Rule

E2 = {H,E , L.O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.E3 = {B,A,D}.P(E3) = 0.9 + 7.3 + 4.4% = 12.6%.E2 and E3 are disjoint events.P(E2 ∪ E3) = P(E2) + P(E3) = 27.4 + 12.6%.

E2 ∪ E3 means “a letter in HELLO or BAD.”E2 ∪ E3 = {H,E , L,O,B,A,D}.Union (∪) corresponds to “or’.’

Math 1710Class 4

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Addition Rule

E2 = {H,E , L.O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.E3 = {B,A,D}.P(E3) = 0.9 + 7.3 + 4.4% = 12.6%.E2 and E3 are disjoint events.P(E2 ∪ E3) = P(E2) + P(E3) = 27.4 + 12.6%.E2 ∪ E3 means “a letter in HELLO or BAD.”

E2 ∪ E3 = {H,E , L,O,B,A,D}.Union (∪) corresponds to “or’.’

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Addition Rule

E2 = {H,E , L.O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.E3 = {B,A,D}.P(E3) = 0.9 + 7.3 + 4.4% = 12.6%.E2 and E3 are disjoint events.P(E2 ∪ E3) = P(E2) + P(E3) = 27.4 + 12.6%.E2 ∪ E3 means “a letter in HELLO or BAD.”E2 ∪ E3 = {H,E , L,O,B,A,D}.

Union (∪) corresponds to “or’.’

Math 1710Class 4

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Addition Rule

E2 = {H,E , L.O}.P(E2) = 3.5 + 13 + 3.5 + 7.4% = 27.4%.E3 = {B,A,D}.P(E3) = 0.9 + 7.3 + 4.4% = 12.6%.E2 and E3 are disjoint events.P(E2 ∪ E3) = P(E2) + P(E3) = 27.4 + 12.6%.E2 ∪ E3 means “a letter in HELLO or BAD.”E2 ∪ E3 = {H,E , L,O,B,A,D}.Union (∪) corresponds to “or’.’

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Equally Likely Outcomes

Sometimes all n outcomes in a sample space are equally likely.

e.g. fair die, fair coins, . . .If so, each has probability 1

n .So the sample space S = {HH,HT ,TH,TT} for tossing 2coins shows us

Num Heads Prob.0 1

41 1

22 1

4

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Sometimes all n outcomes in a sample space are equally likely.e.g. fair die, fair coins, . . .

If so, each has probability 1n .

So the sample space S = {HH,HT ,TH,TT} for tossing 2coins shows us

Num Heads Prob.0 1

41 1

22 1

4

Math 1710Class 4

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Sometimes all n outcomes in a sample space are equally likely.e.g. fair die, fair coins, . . .If so, each has probability 1

n .

So the sample space S = {HH,HT ,TH,TT} for tossing 2coins shows us

Num Heads Prob.0 1

41 1

22 1

4

Math 1710Class 4

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Sometimes all n outcomes in a sample space are equally likely.e.g. fair die, fair coins, . . .If so, each has probability 1

n .So the sample space S = {HH,HT ,TH,TT} for tossing 2coins shows us

Num Heads Prob.0 1

41 1

22 1

4

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Independence Informally

Independence of A and B means knowing A happened doesn’taffect the chance of B happening.

Product Rule for independent events: P(A ∩ B) = P(A)P(B).Intersection (∩) corresponds to “and’.

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Independence of A and B means knowing A happened doesn’taffect the chance of B happening.Product Rule for independent events: P(A ∩ B) = P(A)P(B).

Intersection (∩) corresponds to “and’.

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Independence of A and B means knowing A happened doesn’taffect the chance of B happening.Product Rule for independent events: P(A ∩ B) = P(A)P(B).Intersection (∩) corresponds to “and’.

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Spinner Prob Asst

Discount Probability10% 6/12=.520% 3/12=.2540% 2/12=.17

100% 1/12=.08

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Customers Spin a Wheel

12 equal size areas.Wheel gives customer discount in %.

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Spinner Prob Asst


100% 1/12=.08

(1) Prob of at least a 40% discount?

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Spinner Prob Asst


100% 1/12=.08

(2) Prob two customers in a row get 10% discount?

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Spinner Prob Asst


100% 1/12=.08

(3) Prob 3 consecutive customers get 20%?

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Spinner Prob Asst


100% 1/12=.08

(4) Prob none of first four get over 20%?

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Spinner Prob Asst


100% 1/12=.08

(5) Prob that first 100% is fifth customer?

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Spinner Prob Asst


100% 1/12=.08

(6) Chance that at least one of the first six shoppers gets lessthan a 100% discount?

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Spinner Prob Asst


100% 1/12=.08(1) Prob of at least a 40% discount?

.17 + .08 = .25(2) Prob two customers in a row get 10% discount?

.5 .5.52 = .25.

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Spinner Prob Asst


100% 1/12=.08(1) Prob of at least a 40% discount?.17 + .08 = .25

(2) Prob two customers in a row get 10% discount?

.5 .5.52 = .25.

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Spinner Prob Asst


100% 1/12=.08(1) Prob of at least a 40% discount?.17 + .08 = .25(2) Prob two customers in a row get 10% discount?

.5 .5.52 = .25.

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Spinner Prob Asst



.5 .5

.52 = .25.

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Spinner Prob Asst



.5 .5.52 = .25.

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Venn Diagram Picture

(No longer assuming A and B are disjoint events.)

31 2

A = {1, 2} B = {2, 3}A ∩ B = {2} A ∪ B = {1, 2, 3}

P(A ∪ B) =P(A) + P(B)− P(A ∩ B)

1, 2, 3 1, 2 2, 3 2

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Playing Sports

Suppose 30% play soccer, 20% basketball, 12% both.Probability of playing at least one?

Soln: .30 + .20− .12 = .38.or

.62

S B.12.18 .08

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Playing Sports

Suppose 30% play soccer, 20% basketball, 12% both.Probability of playing at least one?Soln: .30 + .20− .12 = .38.

or

.62

S B.12.18 .08

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Playing Sports

Suppose 30% play soccer, 20% basketball, 12% both.Probability of playing at least one?Soln: .30 + .20− .12 = .38.or

.62

S B.12.18 .08

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Ace or Red

One card is drawn.Probability of an ace or red card?

Direct counting possible:26 red cards + 2 black aces out of 52.Inclusion/Exclusion:P(Ace or Red) = P(Ace) +P(Red) - P(Ace and Red)

=4

52+

26

52− 2

52=

28

52

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Ace or Red

One card is drawn.Probability of an ace or red card?Direct counting possible:26 red cards + 2 black aces out of 52.

Inclusion/Exclusion:P(Ace or Red) = P(Ace) +P(Red) - P(Ace and Red)

=4

52+

26

52− 2

52=

28

52

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Ace or Red

One card is drawn.Probability of an ace or red card?Direct counting possible:26 red cards + 2 black aces out of 52.Inclusion/Exclusion:P(Ace or Red) = P(Ace) +P(Red) - P(Ace and Red)

=4

52+

26

52− 2

52=

28

52

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Smallest Card a Queen

Two cards dealt.Prob(smallest a queen)?(Q viewed as smallest when two Q’s.Ace viewed as high.)

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Two cards dealt.Prob(smallest a queen)?

Approach:

A = event 1st card =Q & 2nd a Q,K or A.

B = event 2nd card =Q & 1st a Q,K or A.

P(A or B) wanted!

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Approach:



P(A or B)=P(A)+P(B)-P(A and B)

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P(A) =4

52· 11

51=

44

51 · 52

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Q

event B

>=Q

4/5211/51

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Q

event B

>=Q

4/5211/51P(B) =

44

51 · 52too.

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P(A and B) =4

52· 3

51=

12

51 · 52

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P(A or B)=P(A)+P(B)-P(A and B)

So P(A or B) =44

51 · 52+

44

51 · 52− 12

51 · 52=

76

51 · 52

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Definition

”Probability of A given B.”

P(A|B) =P(A ∩ B)

P(B)

31 2

A = {1, 2} B = {2, 3}A ∩ B = {2}

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Life Tables

Really a table of probabilities;e.g. P(woman lives to 45)=.963.Will write as P(45).

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Life Tables

P(80|60)?

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Life Tables

P(80|60) =P(80 ∩ 60)

P(60)

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Life Tables

P(80|60) =P(80 ∩ 60)

P(60)=

P(80)

P(60)=.571

.898= .636

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Life Tables

P(80|60) =P(80 ∩ 60)

P(60)=

P(80)

P(60)=.571

.898= .636

P(60|80) =P(80 ∩ 60)

P(80)

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Life Tables

P(80|60) =P(80 ∩ 60)

P(60)=

P(80)

P(60)=.571

.898= .636

P(60|80) =P(80 ∩ 60)

P(80)=

P(80)

P(80)= 1

Math 1710Class 4

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Life Tables

P(80|60) =P(80 ∩ 60)

P(60)=

P(80)

P(60)=.571

.898= .636

P(60|80) =P(80 ∩ 60)

P(80)=

P(80)

P(80)= 1

P(60|80) 6= P(80|60).

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Independence

A and B are independent events if

P(A|B) = P(A)

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Independence


P(A|B) = P(A)

Suppose 10% of all cars have sticky valves10% have oil leaks1% both.Are sticky valves and oil leaks independent?

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Independence


P(A|B) = P(A)

Suppose 10% of all cars have sticky valves10% have oil leaks1% both.Are sticky valves and oil leaks independent?

P(sticky |leaky) =P(both)

P(leaky)=.01

.10= .10 = P(sticky).

So they are independent.

Math 1710Class 4

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Independence


P(A|B) = P(A)

Box has 2 white tickets labeled 12 black labeled 11 white labeled 81 black one labeled 6.Are drawing a 1 and color black independent?

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Independence


P(A|B) = P(A)

Box has 2 white tickets labeled 12 black labeled 11 white labeled 81 black one labeled 6.Are drawing a 1 and color black independent?

P(1|black) =2

3P(1) =

4

6Equal, so they are independent.

Math 1710Class 4

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Independence


P(A|B) = P(A)

Box has 2 white tickets labeled 12 black labeled 11 white labeled 81 black one labeled 6.Are drawing an 8 and color black independent?

Math 1710Class 4

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Misapplications

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Independence


P(A|B) = P(A)

Box has 2 white tickets labeled 12 black labeled 11 white labeled 81 black one labeled 6.Are drawing an 8 and color black independent?

P(8|black) =0

3P(8) =

1

6Not equal, so not independent.

Math 1710Class 4

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Collins Case

Purse Snatching Suspects:

Yellow Auto: 1/10Man with Mustache: 1/4Woman with Ponytail: 1/10Woman with Blond Hair: 1/3Black Man with Beard: 1/10Interracial Couple in Car: 1/1000

Math Prof in LA:1/12,000,000 chance any couple matches these.

Defendants convicted.

Reversed on appeal.

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Misapplications

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Prosecutor’s Fallacy

Just because

P(DNA evidence | innocent)

is small, it doesn’t follow that

P(innocent | DNA evidence)

is small.

“The chance that anyone else but the appellant left the hairsat the scene of the crime is 6 billion to 1.” (1991)

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Defendant’s Fallacy

1 in a million match this DNA

So there are hundreds of matchesand only a small chance the defendant is guilty.

Math 1710Class 4

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From a 2 way table

Jeans OtherMale 12 5

Female 8 11

Math 1710Class 4

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From a 2 way table

With Marginal Distributions:Jeans Other

Male 12 5 17Female 8 11 19

20 16 36

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(M) ?

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(M) ?

P(M) =17

36

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(Male wears jeans)?

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(Male wears jeans)?i.e. P(J|M)

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(J|M) =P(J ∩M)

P(M)

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(J|M) =P(J ∩M)

P(M)=

12

17

More precisely:12

36/

17

36

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(Someone wearing jeans is male)?

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

P(Someone wearing jeans is male)?i.e. P(M|J)

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(M|J) =P(J ∩M)

P(J)

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(M|J) =P(J ∩M)

P(J)=

12

20

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36

Are gender and wearing jeans disjoint?independent?

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(M ∩ J) > 0

So they sometimes both happen.Not disjoint.

Math 1710Class 4

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From a 2 way table


Male 12 5 17Female 8 11 19

20 16 36


P(M|J) 6= P(M)

So not independent.

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Most Pairs of Events are Neither!

Checking disjointness doesn’t even require probabilities.Checking independence does.But when events are related in one of these ways, it is worthnoting.

Math 1710Class 4

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Checking disjointness doesn’t even require probabilities.Checking independence does.But when events are related in one of these ways, it is worthnoting.Sps:A=getting swine fluB=getting a shot.

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Checking disjointness doesn’t even require probabilities.Checking independence does.But when events are related in one of these ways, it is worthnoting.Sps:A=getting swine fluB=getting a shot.What would disjoint/independent mean here?

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Disjointness and Independence Never HappenTogether (almost)

This holds as long as none of the events have probability 0.

For if A and B are disjoint,

P(A|B) =P(A ∩ B)

P(B)= 0 6= P(A).

So A and B are not independent.

(Events of probability 0 do sometimes happen; e.g. a personexactly 173 cm. tall.)

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A Simple Sequential Example

Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?

Math 1710Class 4

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Independence

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.

Math 1710Class 4

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Independence

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).

Math 1710Class 4

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FormalProbability

Letters

IndependentEvents

Spinner

Gen. AdditionRule



Independence

Misapplications

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).First Step:

notA

A

A

Math 1710Class 4

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FormalProbability

Letters

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Independence

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).Second Step:

notA

A

notB

notBB

A B

B

Math 1710Class 4

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).Full First Step:

P(not A)

P(A)

A

A

not A

Math 1710Class 4

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Letters

IndependentEvents

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Independence

Misapplications

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).Full Second Step:

P(not A)

A

P(B|A)

B

P(notB|A)

P(B|not A)

P(notB|notA)

P(A)A

notA

not BB

not B

B

Math 1710Class 4

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Letters

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Independence

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Sequences of events

Tree diagrams are good for keeping track of all thecombinations in multi-stage events.Graphical version of the mult rule P(A ∩ B) = P(B|A)P(A).Final After Multiplying:

P(not A)

A

P(B|A)

B

P(notB|A)

P(B|not A)

P(notB|notA)

P(A and B)

P(A and not B)

P(not B and not A)

P(A)A

notA

not BB

not B

B

P(B and not A)

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Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?

Math 1710Class 4

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Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?Solution: A=win 1st, B=win 2nd

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Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?First Step:

wn1

ls1

Math 1710Class 4

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Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?Second Step:

wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

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Two games against same opponent.P(win 1st)=.4If you win 1st, P(win 2nd)=.2If you lose 1st, P(win 2nd)=.3P(win both)=?; P(win exactly 1 of 2) = ?Probabilities:

wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

.4

.6

.2

.8

.3

.7

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wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

.4

.6

.2

.8

.3

.7

(.4)(.2)=.08

(.4)(.8)=.32

(.6)(.3)=.18

(.6)(.7)=.42

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wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

.4

.6

.2

.8

.3

.7

(.4)(.2)=.08

(.4)(.8)=.32

(.6)(.3)=.18

(.6)(.7)=.42

Note: .08+.32+.18+.42=1since they count allthe disjoint possibilitiesof 1st game/2nd game.

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wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

.4

.6

.2

.8

.3

.7

(.4)(.2)=.08

(.4)(.8)=.32

(.6)(.3)=.18

(.6)(.7)=.42


P(wn 1 of 2)=.32+.18=.50

P(wn 2nd) = .08+.18=.26

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wn1

ls1

wn2

ls2

ls2

1st game 2nd game

wn2

.4

.6

.2

.8

.3

.7

(.4)(.2)=.08

(.4)(.8)=.32

(.6)(.3)=.18

(.6)(.7)=.42


P(wn 1 of 2)=.32+.18=.50

P(wn 2nd) = .08+.18=.26

P(wn 1st|wn 2nd)= P(wn1st&wn2nd)/P(wn2nd) =.08/.26=.308

Math 1710 Class 4pi.math.cornell.edu/~web1710/slides/aug31_v1.pdf · Math 1710 Class 4 V1 Formal...

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Transcript of Math 1710 Class 4pi.math.cornell.edu/~web1710/slides/aug31_v1.pdf · Math 1710 Class 4 V1 Formal...