More ProbabilityMathematics 15: Lecture 14

Dan Sloughter

Furman University

October 17, 2006

Dan Sloughter (Furman University) More Probability October 17, 2006 1 / 14

Laplace’s fifth principle

I Given events E and F ,

P(E |F ) =P(EF )

P(F ).

I Note: this follows from the fourth principle:

P(EF ) = P(E |F )P(F ).

I Note: if E and F are independent

P(E |F ) =P(EF )

P(F )=

P(E )P(F )

P(F )= P(E ).

I Laplace thought of F as an event in the past, which may influencethe probability of a future event E .

I That is, if P(E |F ) differs from P(E ), then F has some influence on,or is a cause of, E .

Dan Sloughter (Furman University) More Probability October 17, 2006 2 / 14

Laplace’s fifth principle

I Given events E and F ,

P(E |F ) =P(EF )

P(F ).

I Note: this follows from the fourth principle:

P(EF ) = P(E |F )P(F ).

I Note: if E and F are independent

P(E |F ) =P(EF )

P(F )=

P(E )P(F )

P(F )= P(E ).

I Laplace thought of F as an event in the past, which may influencethe probability of a future event E .

I That is, if P(E |F ) differs from P(E ), then F has some influence on,or is a cause of, E .

Dan Sloughter (Furman University) More Probability October 17, 2006 2 / 14

Laplace’s fifth principle

I Given events E and F ,

P(E |F ) =P(EF )

P(F ).

I Note: this follows from the fourth principle:

P(EF ) = P(E |F )P(F ).

I Note: if E and F are independent

P(E |F ) =P(EF )

P(F )=

P(E )P(F )

P(F )= P(E ).

I Laplace thought of F as an event in the past, which may influencethe probability of a future event E .

I That is, if P(E |F ) differs from P(E ), then F has some influence on,or is a cause of, E .

Dan Sloughter (Furman University) More Probability October 17, 2006 2 / 14

Laplace’s fifth principle

I Given events E and F ,

P(E |F ) =P(EF )

P(F ).

I Note: this follows from the fourth principle:

P(EF ) = P(E |F )P(F ).

I Note: if E and F are independent

P(E |F ) =P(EF )

P(F )=

P(E )P(F )

P(F )= P(E ).

I Laplace thought of F as an event in the past, which may influencethe probability of a future event E .

I That is, if P(E |F ) differs from P(E ), then F has some influence on,or is a cause of, E .

Dan Sloughter (Furman University) More Probability October 17, 2006 2 / 14

Laplace’s fifth principle

I Given events E and F ,

P(E |F ) =P(EF )

P(F ).

I Note: this follows from the fourth principle:

P(EF ) = P(E |F )P(F ).

I Note: if E and F are independent

P(E |F ) =P(EF )

P(F )=

P(E )P(F )

P(F )= P(E ).

I Laplace thought of F as an event in the past, which may influencethe probability of a future event E .

I That is, if P(E |F ) differs from P(E ), then F has some influence on,or is a cause of, E .

Dan Sloughter (Furman University) More Probability October 17, 2006 2 / 14

Laplace’s seventh principle

I For any event E , let E c denote the event that E does not occur. Thisis the complementary event.

I Example: if E is the event that we obtain at least two heads in threetosses of a coin, then E c is the event we obtain less than two heads,that is, at most one head, in the three tosses.

I Note: we must have P(E c) = 1 − P(E ).

I In its simplest form, Laplace’s seventh principle says: if E and F areany events, then

P(E ) = P(EF ) + P(EF c) = P(E |F )P(F ) + P(E |F c)P(F c).

Dan Sloughter (Furman University) More Probability October 17, 2006 3 / 14

Laplace’s seventh principle

I For any event E , let E c denote the event that E does not occur. Thisis the complementary event.

I Example: if E is the event that we obtain at least two heads in threetosses of a coin, then E c is the event we obtain less than two heads,that is, at most one head, in the three tosses.

I Note: we must have P(E c) = 1 − P(E ).

I In its simplest form, Laplace’s seventh principle says: if E and F areany events, then

P(E ) = P(EF ) + P(EF c) = P(E |F )P(F ) + P(E |F c)P(F c).

Dan Sloughter (Furman University) More Probability October 17, 2006 3 / 14

Laplace’s seventh principle

I For any event E , let E c denote the event that E does not occur. Thisis the complementary event.

I Example: if E is the event that we obtain at least two heads in threetosses of a coin, then E c is the event we obtain less than two heads,that is, at most one head, in the three tosses.

I Note: we must have P(E c) = 1 − P(E ).

I In its simplest form, Laplace’s seventh principle says: if E and F areany events, then

P(E ) = P(EF ) + P(EF c) = P(E |F )P(F ) + P(E |F c)P(F c).

Dan Sloughter (Furman University) More Probability October 17, 2006 3 / 14

Laplace’s seventh principle

I For any event E , let E c denote the event that E does not occur. Thisis the complementary event.

I Example: if E is the event that we obtain at least two heads in threetosses of a coin, then E c is the event we obtain less than two heads,that is, at most one head, in the three tosses.

I Note: we must have P(E c) = 1 − P(E ).

I In its simplest form, Laplace’s seventh principle says: if E and F areany events, then

P(E ) = P(EF ) + P(EF c) = P(E |F )P(F ) + P(E |F c)P(F c).

Dan Sloughter (Furman University) More Probability October 17, 2006 3 / 14

Example

I Suppose we have two urns, one which has two white balls and onewhich has one white and one black ball. We choose an urn and selecta ball from the urn.

I Let E be the event that the selected ball is white.

I If we let F be the event we selected the urn with two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

2× 1

2=

3

4.

I Now suppose we replace the first ball and draw a second, from thesame urn. If E is the event we draw two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

4× 1

2=

5

8.

Dan Sloughter (Furman University) More Probability October 17, 2006 4 / 14

Example

I Suppose we have two urns, one which has two white balls and onewhich has one white and one black ball. We choose an urn and selecta ball from the urn.

I Let E be the event that the selected ball is white.

I If we let F be the event we selected the urn with two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

2× 1

2=

3

4.

I Now suppose we replace the first ball and draw a second, from thesame urn. If E is the event we draw two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

4× 1

2=

5

8.

Dan Sloughter (Furman University) More Probability October 17, 2006 4 / 14

Example

I Suppose we have two urns, one which has two white balls and onewhich has one white and one black ball. We choose an urn and selecta ball from the urn.

I Let E be the event that the selected ball is white.

I If we let F be the event we selected the urn with two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

2× 1

2=

3

4.

I Now suppose we replace the first ball and draw a second, from thesame urn. If E is the event we draw two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

4× 1

2=

5

8.

Dan Sloughter (Furman University) More Probability October 17, 2006 4 / 14

Example

I Suppose we have two urns, one which has two white balls and onewhich has one white and one black ball. We choose an urn and selecta ball from the urn.

I Let E be the event that the selected ball is white.

I If we let F be the event we selected the urn with two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

2× 1

2=

3

4.

I Now suppose we replace the first ball and draw a second, from thesame urn. If E is the event we draw two white balls, then

P(E ) = P(E |F )P(F ) + P(E |F c)P(F c) = 1 × 1

2+

1

4× 1

2=

5

8.

Dan Sloughter (Furman University) More Probability October 17, 2006 4 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:

I Suppose we have 4 urns, each containing three balls, with three, two,one, and zero white balls.

I Experiment: pick an urn, draw out a ball, replace the ball, and thendraw out a second ball.

I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:

I Suppose we have 4 urns, each containing three balls, with three, two,one, and zero white balls.

I Experiment: pick an urn, draw out a ball, replace the ball, and thendraw out a second ball.

I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:

I Suppose we have 4 urns, each containing three balls, with three, two,one, and zero white balls.

I Experiment: pick an urn, draw out a ball, replace the ball, and thendraw out a second ball.

I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:I Suppose we have 4 urns, each containing three balls, with three, two,

one, and zero white balls.

I Experiment: pick an urn, draw out a ball, replace the ball, and thendraw out a second ball.

I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:I Suppose we have 4 urns, each containing three balls, with three, two,

one, and zero white balls.I Experiment: pick an urn, draw out a ball, replace the ball, and then

draw out a second ball.

I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

More general statement

I More generally, suppose exactly one of F1, F2, . . . , Fn must haveoccurred.

I Then

P(E ) = P(E |F1)P(F1) + P(E |F2)P(F2) + · · · + P(E |Fn)P(Fn).

I Example:I Suppose we have 4 urns, each containing three balls, with three, two,

one, and zero white balls.I Experiment: pick an urn, draw out a ball, replace the ball, and then

draw out a second ball.I If E is the event that both balls are white, then

P(E ) = 1 × 1

4+

4

9× 1

4+

1

9× 1

4+ 0 × 1

4=

14

36=

7

18.

Dan Sloughter (Furman University) More Probability October 17, 2006 5 / 14

Laplace’s sixth principle

I For any two events E and F ,

P(F |E ) =P(FE )

P(E )=

P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c).

I This is Bayes’ Theorem, named for Thomas Bayes (1702 - 1761).

Dan Sloughter (Furman University) More Probability October 17, 2006 6 / 14

Laplace’s sixth principle

I For any two events E and F ,

P(F |E ) =P(FE )

P(E )=

P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c).

I This is Bayes’ Theorem, named for Thomas Bayes (1702 - 1761).

Dan Sloughter (Furman University) More Probability October 17, 2006 6 / 14

Example

I Suppose we have two urns, one with two white balls and one with onewhite and one red ball.

I We pick an urn at random and draw one ball. If the ball is white, whatis the probability we are choosing from the urn with two white balls?

I Let F be the event we are choosing from the urn with two white ballsand let E be the event that we draw a white ball.

I Then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)=

1 × 12

1 × 12 + 1

2 ×12

=2

3.

I Note: a priori, that is, prior to any evidence, the probability that weare drawing from the urn with two white balls is 1

2 . After seeing theevidence, that is, that we drew a white ball, the probability goes up to23 .

Dan Sloughter (Furman University) More Probability October 17, 2006 7 / 14

Example

I Suppose we have two urns, one with two white balls and one with onewhite and one red ball.

I We pick an urn at random and draw one ball. If the ball is white, whatis the probability we are choosing from the urn with two white balls?

I Let F be the event we are choosing from the urn with two white ballsand let E be the event that we draw a white ball.

I Then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)=

1 × 12

1 × 12 + 1

2 ×12

=2

3.

I Note: a priori, that is, prior to any evidence, the probability that weare drawing from the urn with two white balls is 1

2 . After seeing theevidence, that is, that we drew a white ball, the probability goes up to23 .

Dan Sloughter (Furman University) More Probability October 17, 2006 7 / 14

Example

I Suppose we have two urns, one with two white balls and one with onewhite and one red ball.

I We pick an urn at random and draw one ball. If the ball is white, whatis the probability we are choosing from the urn with two white balls?

I Let F be the event we are choosing from the urn with two white ballsand let E be the event that we draw a white ball.

I Then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)=

1 × 12

1 × 12 + 1

2 ×12

=2

3.

I Note: a priori, that is, prior to any evidence, the probability that weare drawing from the urn with two white balls is 1

2 . After seeing theevidence, that is, that we drew a white ball, the probability goes up to23 .

Dan Sloughter (Furman University) More Probability October 17, 2006 7 / 14

Example

I Suppose we have two urns, one with two white balls and one with onewhite and one red ball.

I We pick an urn at random and draw one ball. If the ball is white, whatis the probability we are choosing from the urn with two white balls?

I Let F be the event we are choosing from the urn with two white ballsand let E be the event that we draw a white ball.

I Then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)=

1 × 12

1 × 12 + 1

2 ×12

=2

3.

I Note: a priori, that is, prior to any evidence, the probability that weare drawing from the urn with two white balls is 1

2 . After seeing theevidence, that is, that we drew a white ball, the probability goes up to23 .

Dan Sloughter (Furman University) More Probability October 17, 2006 7 / 14

Example

I Suppose we have two urns, one with two white balls and one with onewhite and one red ball.

I We pick an urn at random and draw one ball. If the ball is white, whatis the probability we are choosing from the urn with two white balls?

I Let F be the event we are choosing from the urn with two white ballsand let E be the event that we draw a white ball.

I Then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)=

1 × 12

1 × 12 + 1

2 ×12

=2

3.

I Note: a priori, that is, prior to any evidence, the probability that weare drawing from the urn with two white balls is 1

2 . After seeing theevidence, that is, that we drew a white ball, the probability goes up to23 .

Dan Sloughter (Furman University) More Probability October 17, 2006 7 / 14

Example

I Suppose a medical test correctly identifies a certain condition 99% ofthe time if the person being tested has the condition and correctlysays a person does not have the condition 98% of the time when theperson does not have the condition.

I Suppose that one out of every 10, 000 people have this condition.

I If E is the event the test is positive and F is the event that a personhas this condition, then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)

=(0.99)(0.0001)

(0.99)(0.0001) + (0.02)(0.9999)= 0.004926.

I Note: the person is 50 times as likely to have condition, but theprobability of having it is still very small.

Dan Sloughter (Furman University) More Probability October 17, 2006 8 / 14

Example

I Suppose a medical test correctly identifies a certain condition 99% ofthe time if the person being tested has the condition and correctlysays a person does not have the condition 98% of the time when theperson does not have the condition.

I Suppose that one out of every 10, 000 people have this condition.

I If E is the event the test is positive and F is the event that a personhas this condition, then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)

=(0.99)(0.0001)

(0.99)(0.0001) + (0.02)(0.9999)= 0.004926.

I Note: the person is 50 times as likely to have condition, but theprobability of having it is still very small.

Dan Sloughter (Furman University) More Probability October 17, 2006 8 / 14

Example

I Suppose a medical test correctly identifies a certain condition 99% ofthe time if the person being tested has the condition and correctlysays a person does not have the condition 98% of the time when theperson does not have the condition.

I Suppose that one out of every 10, 000 people have this condition.

I If E is the event the test is positive and F is the event that a personhas this condition, then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)

=(0.99)(0.0001)

(0.99)(0.0001) + (0.02)(0.9999)= 0.004926.

I Note: the person is 50 times as likely to have condition, but theprobability of having it is still very small.

Dan Sloughter (Furman University) More Probability October 17, 2006 8 / 14

Example

I Suppose a medical test correctly identifies a certain condition 99% ofthe time if the person being tested has the condition and correctlysays a person does not have the condition 98% of the time when theperson does not have the condition.

I Suppose that one out of every 10, 000 people have this condition.

I If E is the event the test is positive and F is the event that a personhas this condition, then

P(F |E ) =P(E |F )P(F )

P(E |F )P(F ) + P(E |F c)P(F c)

=(0.99)(0.0001)

(0.99)(0.0001) + (0.02)(0.9999)= 0.004926.

I Note: the person is 50 times as likely to have condition, but theprobability of having it is still very small.

Dan Sloughter (Furman University) More Probability October 17, 2006 8 / 14

Laplace’s example on page 1332

I Suppose we have three urns: Urn I has two white balls, Urn II has onewhite and one black ball, and Urn III has two black balls.

I We pick an urn at random and start sampling balls from the urn,replacing a ball after each draw.

I Question: If the first two balls sampled are white, what is theprobability that the third ball drawn will be white?

Dan Sloughter (Furman University) More Probability October 17, 2006 9 / 14

Laplace’s example on page 1332

I Suppose we have three urns: Urn I has two white balls, Urn II has onewhite and one black ball, and Urn III has two black balls.

I We pick an urn at random and start sampling balls from the urn,replacing a ball after each draw.

I Question: If the first two balls sampled are white, what is theprobability that the third ball drawn will be white?

Dan Sloughter (Furman University) More Probability October 17, 2006 9 / 14

Laplace’s example on page 1332

I Suppose we have three urns: Urn I has two white balls, Urn II has onewhite and one black ball, and Urn III has two black balls.

I We pick an urn at random and start sampling balls from the urn,replacing a ball after each draw.

I Question: If the first two balls sampled are white, what is theprobability that the third ball drawn will be white?

Dan Sloughter (Furman University) More Probability October 17, 2006 9 / 14

Laplace’s example (cont’d)

I By Baye’s Theorem, given that we have sampled two white balls, theprobability that we are sampling from

I Urn I is1 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

=4

5,

I Urn II is14 ×

13

1 × 13 + 1

4 ×13 + 0 × 1

3

=1

5,

I Urn III is0 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

= 0.

I Hence the probability that the next ball drawn is white is

1 × 4

5+

1

2× 1

5+ 0 × 0 =

9

10.

Dan Sloughter (Furman University) More Probability October 17, 2006 10 / 14

Laplace’s example (cont’d)

I By Baye’s Theorem, given that we have sampled two white balls, theprobability that we are sampling from

I Urn I is1 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

=4

5,

I Urn II is14 ×

13

1 × 13 + 1

4 ×13 + 0 × 1

3

=1

5,

I Urn III is0 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

= 0.

I Hence the probability that the next ball drawn is white is

1 × 4

5+

1

2× 1

5+ 0 × 0 =

9

10.

Dan Sloughter (Furman University) More Probability October 17, 2006 10 / 14

Laplace’s example (cont’d)

I By Baye’s Theorem, given that we have sampled two white balls, theprobability that we are sampling from

I Urn I is1 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

=4

5,

I Urn II is14 ×

13

1 × 13 + 1

4 ×13 + 0 × 1

3

=1

5,

I Urn III is0 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

= 0.

I Hence the probability that the next ball drawn is white is

1 × 4

5+

1

2× 1

5+ 0 × 0 =

9

10.

Dan Sloughter (Furman University) More Probability October 17, 2006 10 / 14

Laplace’s example (cont’d)

I By Baye’s Theorem, given that we have sampled two white balls, theprobability that we are sampling from

I Urn I is1 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

=4

5,

I Urn II is14 ×

13

1 × 13 + 1

4 ×13 + 0 × 1

3

=1

5,

I Urn III is0 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

= 0.

I Hence the probability that the next ball drawn is white is

1 × 4

5+

1

2× 1

5+ 0 × 0 =

9

10.

Dan Sloughter (Furman University) More Probability October 17, 2006 10 / 14

Laplace’s example (cont’d)

I By Baye’s Theorem, given that we have sampled two white balls, theprobability that we are sampling from

I Urn I is1 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

=4

5,

I Urn II is14 ×

13

1 × 13 + 1

4 ×13 + 0 × 1

3

=1

5,

I Urn III is0 × 1

3

1 × 13 + 1

4 ×13 + 0 × 1

3

= 0.

I Hence the probability that the next ball drawn is white is

1 × 4

5+

1

2× 1

5+ 0 × 0 =

9

10.

Dan Sloughter (Furman University) More Probability October 17, 2006 10 / 14

The law of succession

I Suppose a coin lands head with probability p, where p is equally likelyto be any number between 0 and 1.

I Suppose we toss the coin n times, each time resulting in a head.

I Laplace calculates that the probability that we will get a head on thenext toss of the coin is n+1

n+2 .

I Assuming the earth is 5000 years old and there are 365.2426 days in ayear, then the sun has risen 1, 826, 213 consecutive days.

I Hence the probability that the sun will rise tomorrow is

1, 826, 214

1, 826, 215= 0.999999452419.

Dan Sloughter (Furman University) More Probability October 17, 2006 11 / 14

The law of succession

I Suppose a coin lands head with probability p, where p is equally likelyto be any number between 0 and 1.

I Suppose we toss the coin n times, each time resulting in a head.

I Laplace calculates that the probability that we will get a head on thenext toss of the coin is n+1

n+2 .

I Assuming the earth is 5000 years old and there are 365.2426 days in ayear, then the sun has risen 1, 826, 213 consecutive days.

I Hence the probability that the sun will rise tomorrow is

1, 826, 214

1, 826, 215= 0.999999452419.

Dan Sloughter (Furman University) More Probability October 17, 2006 11 / 14

The law of succession

I Suppose a coin lands head with probability p, where p is equally likelyto be any number between 0 and 1.

I Suppose we toss the coin n times, each time resulting in a head.

I Laplace calculates that the probability that we will get a head on thenext toss of the coin is n+1

n+2 .

I Assuming the earth is 5000 years old and there are 365.2426 days in ayear, then the sun has risen 1, 826, 213 consecutive days.

I Hence the probability that the sun will rise tomorrow is

1, 826, 214

1, 826, 215= 0.999999452419.

Dan Sloughter (Furman University) More Probability October 17, 2006 11 / 14

The law of succession

I Suppose a coin lands head with probability p, where p is equally likelyto be any number between 0 and 1.

I Suppose we toss the coin n times, each time resulting in a head.

I Laplace calculates that the probability that we will get a head on thenext toss of the coin is n+1

n+2 .

I Assuming the earth is 5000 years old and there are 365.2426 days in ayear, then the sun has risen 1, 826, 213 consecutive days.

I Hence the probability that the sun will rise tomorrow is

1, 826, 214

1, 826, 215= 0.999999452419.

Dan Sloughter (Furman University) More Probability October 17, 2006 11 / 14

The law of succession

I Suppose a coin lands head with probability p, where p is equally likelyto be any number between 0 and 1.

I Suppose we toss the coin n times, each time resulting in a head.

I Laplace calculates that the probability that we will get a head on thenext toss of the coin is n+1

n+2 .

I Assuming the earth is 5000 years old and there are 365.2426 days in ayear, then the sun has risen 1, 826, 213 consecutive days.

I Hence the probability that the sun will rise tomorrow is

1, 826, 214

1, 826, 215= 0.999999452419.

Dan Sloughter (Furman University) More Probability October 17, 2006 11 / 14

The law of succession (cont’d)

I C. S. Peirce (1839 - 1914): this might make sense, if “universes wereas plenty as blackberries.”

I That is: is our universe just one chosen from an infinite number ofequally likely universes?

I Recall: Laplace would say it’s not that our universe has been chosenat random; rather, probability is a measure of our ignorance ofultimate causes.

Dan Sloughter (Furman University) More Probability October 17, 2006 12 / 14

The law of succession (cont’d)

I C. S. Peirce (1839 - 1914): this might make sense, if “universes wereas plenty as blackberries.”

I That is: is our universe just one chosen from an infinite number ofequally likely universes?

I Recall: Laplace would say it’s not that our universe has been chosenat random; rather, probability is a measure of our ignorance ofultimate causes.

Dan Sloughter (Furman University) More Probability October 17, 2006 12 / 14

The law of succession (cont’d)

I C. S. Peirce (1839 - 1914): this might make sense, if “universes wereas plenty as blackberries.”

I That is: is our universe just one chosen from an infinite number ofequally likely universes?

I Recall: Laplace would say it’s not that our universe has been chosenat random; rather, probability is a measure of our ignorance ofultimate causes.

Dan Sloughter (Furman University) More Probability October 17, 2006 12 / 14

Problems

1. A company stocks widgets from two suppliers. Suppose that 4% ofthe widgets from Supplier I are defective and 8% of the widgets fromSupplier II are defective. If the company buys 65% of its widgets fromSupplier I and 35% of its widgets from Supplier II, what is theprobability that a randomly selected widget from the stock of thecompany will be defective?

2. Suppose Urn I has two white balls and three black balls and Urn IIhas three white balls and one black ball. An urn is chosen at randomand balls are sampled, with replacement, from it.

a. Find the probability that the first ball selected is white.b. Find the probability that the first two balls selected are white.c. If the first ball selected is white, what is the probability that you are

sampling from Urn I?d. If the first two balls selected are white, what is the probability that you

are sampling from Urn I?

Dan Sloughter (Furman University) More Probability October 17, 2006 13 / 14

Problems (cont’d)

3. To determine which of its 100 employees is guilty of stealing, acompany decides to use a lie detector test. Suppose the test correctlyidentifies a person as lying when the person is lying 95% of the timeand correctly identifies that someone is telling the truth when thatperson is telling the truth 90% of the time. If a person given the testdenies that he is the thief, but the lie detector says he is lying, what isthe probability he is the thief?

Dan Sloughter (Furman University) More Probability October 17, 2006 14 / 14