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15.Math-Review

Thursday 8/17/00Thursday 8/17/00

15.Math-Review 2

Venn Diagrams

Event A, and event Ac, the complement of A:

AAc

U

The Universal Set, U

Notation: The complete set:

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Notation: A and B, called AB A or B, called AB

Venn Diagrams

A B

A B

A B

U

A

B

U B included in A, or BA

The null set is labeled

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Axioms1. A B = B A

2. A ( B C ) = ( A B ) C

3. A ( B C ) = ( A B ) ( A C )

4. ( A c) c = A

5. ( A B )c = Ac Bc

6. A Ac = 7. A U = A

Venn Diagrams

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And the relation to probability. Consider the uncertain event of drawing a card out of

a deck of cards:Possible outcomes:

Venn Diagrams

SA SK SQ SJ ST S9 S8 S7 S6 S5 S4 S3 S2

HA HK HQ HJ HT H9 H8 H7 H6 H5 H4 H3 H2

DA DK DQ DJ DT D9 D8 D7 D6 D5 D4 D3 D2

CA CK CQ CJ CT C9 C8 C7 C6 C5 C4 C3 C2

Each of the outcomes is equally likely, and mutually exclusive. With probability 1/52.

We also have uncertain events like:• Draw a face card• Draw a heart.

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Venn Diagrams

SA SK SQ SJ ST S9 S8 S7 S6 S5 S4 S3 S2

HA HK HQ HJ HT H9 H8 H7 H6 H5 H4 H3 H2

DA DK DQ DJ DT D9 D8 D7 D6 D5 D4 D3 D2

CA CK CQ CJ CT C9 C8 C7 C6 C5 C4 C3 C2

These uncertain events can be represented as subsets of the possible outcomes.

In this example, The two events are not mutually exclusive.The probability of each is the sum of the outcomes it contains.

Subsets of the universe of possible outcomes are events.

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Event: In this setting we are talking about some uncertain event. The

outcome of which is uncertain

Outcome: The result of an observation of the event once the uncertainty has

been resolved.

Probability: The likelihood that certain outcome is realized for the event.

Example:To roll a balanced 6-sided die is an event. The number that appears

on top of the die once its rolled it’s the outcome. And any outcome (any of the 6 sides) has a probability of 1/6.

Laws of Probability

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Laws of Probability The probability of any event is a number between 0 and 1:

0 P(A) 1 If A and B are mutually exclusive events, then:

P( A B ) = P(A) + P(B) Two events are mutually exclusive if the occurrence of one totally

prohibits the occurrence of the other. In other words AB=

Laws of Probability

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If A and B are independent, then P( A B ) = P(A)P(B) Two events are independent if news that one of them took place

neither raises nor lowers the chance that the other did so. Independence is very different from mutual exclusivity! If A and

B are independent, the fact that A occurred leaves the probability of B untouched; if A and B are mutually exclusive, the same fact would make the chance of B plunge to 0.

Conditional probability: for any two events A and B, the conditional probability of A given B is:

Laws of Probability

)(

)()|(

BP

BAPBAP

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Exercise: Two fair six-sided dice are tossed. Find the probability that: a. the sum of the outcomes is exactly 2 b. the sum of the outcomes exceeds 2 c. both dice come up with the same number d. both dice emerge with odd numbers e. event (a) occurs given that event (c) does

Probability Exercises

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Exercise: Consider two events about a piece of equipment for which both the primary and backup systems are up and running now:

A1: the primary system fails in the next hour A2: the backup system fails in the next hour Let p1 = P(A1) and p2 = P(A2)

Suppose that p1 and p2 are exactly the same, both of them sharing the common value, p. Suppose also that A1 and A2 are independent events.

Probability Exercises

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Exercise: What is the probability that:

a. both systems fail in the next hour

b. the primary system does not fail over the next hour

c. the primary system fails in the next hour but not the backup

d. exactly one of the two systems fails in the next hour

e. at least one system fails in the next hour

Suppose p1 p2. Under this modification, find the chance that:

f. both systems fail in the next hour

g. exactly one of the systems fails

h. at least one of the systems fails

Probability Exercises

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Exercise: Everyone buying a pizza at JR’s pizzeria gets a card that conceals either a J or an R. When the wax covering is rubbed off, the chance is 80% that a J will appear. Different cards can be treated as having independent letters.

Once a person has both a J and R, she is entitled to a free pizza. Note that, because the same letter can come up on several successive purchases, that rule is not equivalent to one free pizza per two paid ones.

Probability Exercises

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Probability Exercises

Exercise: If Minerva starts patronizing the pizzeria, find the probability that:

a. her first two purchases yield a J and R, in that order

b. she is eligible for a free pizza after her first two purchases

c. it takes her exactly three purchases to achieve the bonus

d. even after buying five pizzas, she is not eligible for her first complimentary one.

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Exercise: Suppose that a study shows that, of the products that recently received patents, 12% later became commercially successful. The study also shows that 20% of recent patents involved biotechnology and that 30% of these biotechnical products were commercially successful.

a. If Mendel had invested in four recently-patented products that he chose at random, what is the probability that all four were commercially successful?

Probability Exercises

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Probability Exercises

Exercise:b. Given that all four of his products were commercially successful, what is

the chance that they all involved biotechnology?

c. Let Q be the fraction of recently patented nonbiotechnological products that were successful. Given the information above, what is the numerical value of Q? (HINT: Express the overall success rate for recently-patented products in terms of Q.)

d. Are the successfulness and biotechnicality of a project independent events?

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Independence

Two events A and B are said to be independent if P(AB) = P(A)P(B).

Or using the definition of conditional probability:

)()(

)()(

)(

)()|( AP

BP

BPAP

BP

BAPBAP

Intuition: The fact that you know the outcome of event B, doesn’t change the probabilities of event A

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Independence

Example: Lets consider the 2 die example. Where we want to obtain a double. In other words a success is going to be if the same number appears in both dice. Remember this problem?

We will call D the event of obtaining a double. A the event of the roll of one die (the first… yes we will distinguish the die). And B the event of the roll of the second die.

All the possible outcomes are the following 36 equally likely combinations: 11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

The probability of each outcome is 1/36

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Independence

)()|(

6/1)6()6|(

...

6/1)2()2|(

6/1)1()1|(

6/1)36/1(6),()(6

1

DPADP

BPADP

BPADP

BPADP

iBiAPDPi

In this example we see that the event of getting a double is independent of the outcome of the first die….

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Independence

If we are conducting this same experiment in Vegas, where its very hard to find a ‘fair’ die, things change a little:

Suppose the actual probability distribution of the 2 Vegas dice we will use are:

xi pi

1 0.1

2 0.2

3 0.3

4 0.2

5 0.1

6 0.1

Now the 36 possible outcomes are no longer equally likely, but we can compute their probability because the roll of the two dice are still independent. For example:

02.01.0*2.0)5()2()5,2( BPAPBAP

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Independence

)()|( have wegeneralin

2.0)2()2|(

1.0)1()1|( and

2.01.01.02.03.02.01.0)(

)()(),()(

2222226

1

2

6

1

6

1

DPADP

BPADP

BPADP

iAP

iBPiAPiBiAPDP

i

ii

In this case the event of drawing a double depends on what we roll with the first die.

Lets compute the conditional probabilities:

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Birthday Problem

Suppose there are N people in a room. What is the probability that no two of the people in the room share the same birthday?

Assumptions:(1) no prior relationships among the birthdays of the people in the room (no twins!); and(2) P(born on any particular day) = 1/365 (i.e. we neglect leap years).

Let’s try it in this room!

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Birthday Problem

Suppose there are N = 50 people in the room. Let Ai be the event the event that i th person states a date that

has not been stated by the previous i - 1 people. Let B be the event that all N people have different birthdays.

Then

P(B) = P(A1 and A2 and ... AN)

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Birthday Problem

Notice that P(A1 and A2 and ... AN) =

P(AN|A1 and A2 and ... AN-1) x P(A1 and A2 and ... AN-1)

Similarly, P(A1 and A2 and ... AN-1) =P(AN-1|A1 and A2 and ... AN-2) x P(A1 and A2 and ... AN-2)

P(A1) = 1, since the first person asked cannot repeat a new date. The second person will announce a new date unless the second person was born the same day as the first person. Therefore

P(A2 | A1) =P(A1,A2)/P(A1) =1(1 - 1/365)/1 = 364/365

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Birthday Problem

To calculate P(A3|A1 and A2), note that if both A1 and A2 have occurred, there are two days the third person must avoid in order to prevent repetition. Therefore,

P(A3|A1,A2)=P(A1,A2,A3)/P(A1,A2)

=1(364/365)(363/365)/(364/365)=363/365

Applying the logic more generally, we obtain that P(Ai|A1 and A2 and ... Ai-1) = [365 - (i - 1)]/365

= (366 - i)/365

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Birthday Problem

P(Ai|A1 and A2 and ... Ai-1) = (366 - i)/365

Then, P(B) = (365) (364) (363) ... (366-N) (365) (365) (365) (365)

So, for N = 50:(365) (364) ... (316) = 0.030(365) (365) (365)

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We have an uncertain event. If the outcome of the uncertainty is a number then it is

called a random variable: Example:

The result of the flip of a coin is uncertain event. We can obtain heads or tails. This is not a random variable.

If we associate the variable X a value equal to 1 if the coin flip is head and 0 if the coin flip is tails, then X is a random variable.

Random Variables

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A random process (or event) is one whose outcome cannot be specified in advance (except in probabilistic terms).

A random variable is a number that reflects the outcome of a random process.

Random Variables

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A random variable can be discrete or continuous. Discrete: The values the random variable can take are fixed

discrete amounts.

Example: The number on top after the roll of a die can be 1, 2, 3, 4, 5 or 6.

Continuous: The random variable can take any value in some interval.

Example: If we draw a student at random in the class and record their height in principle we can obtain any number between .5 meters and 2.5 meters. (Very unlikely in the extremes)

Random Variables