概率与统计入门教程：理论与实战应用

数理统计

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18.05 class 2, Probability: Terminology and Examples, Spring 2014

2.4 The probability function

So far we’ve been using a casual deﬁnition of the probability function. Let’s give a more

precise one.

Careful deﬁnition of the probability function.

For a discrete sample space S a probability function P assigns to each outcome ω anumber

P (ω) called the probability of ω. P must satisfy two rules:

Rule 1. 0 ≤ P (ω) ≤ 1 (probabilities are between 0 and 1).

Rule 2. The sum of the probabilities of all possible outcomes is 1 (something must occur)

In symbols Rule 2 says: if S = {ω

,ω

,...,ω

} then P (ω

)+ P (ω

)+ ...+ P (ω

)=1.

Using summation notation: P (ω

)=1.

j=1

The probability of an event E is the sum of the probabilities of all the outcomes in E.That

is,

P (E)= P (ω).

ω∈E

Think: Check Rules 1 and 2 on Examples 1 and 2 above.

Example 7. Flip until heads (A classic example)

Suppose we have a coin with probability p of heads.

Experiment: Toss the coin until the ﬁrst heads. Report the number of tosses.

Sample space: Ω = {1, 2, 3, . . . }.

Probability function: P (n)=(1 − p)

n−1

Challenge 1: show the sum of all the probabilities equals 1 (hint: geometric series).

Challenge 2: justify the formula for P (n) (we will do this soon).

Stopping problems. The previous toy example is an uncluttered version of a general

class of problems called stopping rule problems. A stopping rule is a rule that tells you

when to end a certain process. In the toy example above the process was ﬂipping a coin

and we stopped after the ﬁrst heads. A more practical example is a rule for ending a series

of medical treatments. Such a rule could depend on how well the treatments are working,

how the patient is tolerating them and the probability that the treatments would continue

to be eﬀective. One could ask about the probability of stopping within a certain number of

treatments or the average number of treatments you should expect before stopping.

Some rules of probability

For events A, L and R contained in a sample space Ω.

Rule 1. P (A

)=1 − P(A).

Rule 2. If L and R are disjoint then P (L ∪ R)= P (L)+ P (R).

Rule 3. If L and R are not disjoint, we have the inclusion-exclusion principle:

P (L ∪ R)= P (L)+ P (R) − P (L ∩ R)

L R L R

5 18.05 class 2, Probability: Terminology and Examples, Spring 2014

We visualize these rules using Venn diagrams.

Ω= A ∪ A ,nooverlap L ∪ R,n

ooverlap L ∪ R, overlap = L ∩ R

We can also justify them logically.

Rule 1: A and A

split Ω into two non-overlapping regions. Since the total probability

P (Ω) = 1 this rule says that the probabiity of A and the probability of ’not A’are comple-

mentary, i.e. sum to 1.

Rule 2: L and R split L ∪ R into two non-overlapping regions. So the probability of L ∪ R

is is split between P (L)and P (R)

Rule 3: In the sum P (L)+ P (R)the overlap P (L ∩ R) gets counted twice. So P (L)+

P (R) − P (L ∩ R) counts everything in the union exactly once.

Think: Rule 2 is a special case of Rule 3.

For the following examples suppose we have an experiment that produces a random integer

between 1 and 20. The probabilities are not necessarily uniform, i.e., not necessarily the

same for each outcome.

Example 8. If the probability of an even number is .6 what is the probability of an odd

number?

answer: Since being odd is complementary to being even, the probability of being odd is

1-.6 = .4.

Let’s redo this example a bit more formally, so you see how it’s done. First, so we can refer

to it, let’s name the random integer X. Let’s also name the event ‘X is even’ as A.Then

the event ‘X is odd’ is A

. We are given that P (A)= .6. Therefore P (A

)=1 − .6= .4 .

Example 9. Consider the 2 events, A:‘X is a multiple of 2’; B:‘X is odd and less than

10’. Suppose P (A)= .6and P (B)= .25.

(i) What is A ∩ B?

(ii) What is the probability of A ∪ B?

answer: (i) Since all numbers in A are even and all numbers in B are odd, these events are

disjoint. That is, A ∩ B = ∅.

(ii) Since A and B are disjoint P (A ∪ B)= P (A)+ P (B)= .85.

Example 10. Let A, B and C be the events X is a multiple of 2, 3 and 6 respectively. If

P (A)= .6, P (B)= .3and P (C)= .2whatis P (A or B)?

answer: Note two things. First we used the word ‘or’ which means union: ‘A or B’=

A ∪ B. Second, an integer is divisible by 6 if and only if it is divisible by both 2 and 3.

This translates into C = A ∩ B. So the inclusion-exclusion principle says

P (A ∪ B)= P (A)+ P (B) − P (A ∩ B)= .6+ .3 − .2= .7 .

Endofclass2notes

Conditional Probability, Independence and Bayes’ Theorem

Class 3, 18.05, Spring 2014

Jeremy Orloﬀ and Jonathan Bloom

1 Learning Goals

1. Know the deﬁnitions of conditional probability and independence of events.

2. Be able to compute conditional probability directly from the deﬁnition.

3. Be able to use the multiplication rule to compute the total probability of an event.

4. Be able to check if two events are independent.

5. Be able to use Bayes formula to ‘invert’ conditional probabilities.

6. Be able to organize the computation of conditional probabilities using trees and tables.

7. Understand the base rate fallacy thoroughly.

2 Conditional Probability

Conditional probability answers the question ‘how does the probability of an event change

if we have extra information’. We’ll illustrate with an example.

Example 1. Toss a fair coin 3 times.

a) What is the probability of 3 heads?

answer: Sample space Ω = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}.

All outcomes are equally likely, so P (3 heads) = 1/8.

b) Suppose we are told that the ﬁrst toss was heads. Given this information how should we

compute the probability of 3 heads?

answer: We have a new (reduced) sample space: Ω

= {HHH,HHT,HTH,HTT}.

All outcomes are equally likely, so

P (3 heads given that the ﬁrst toss is heads) = 1/4.

This is called conditional probability, since it takes into account additional conditions. To

develop the notation, we rephrase (b) in terms of events.

b) Let A be the event ‘all three tosses are heads’ = {HHH}.

Let B be the event ‘the ﬁrst toss is heads’ = {HHH,HHT,HTH,HTT}.

The conditional probability of A knowing that B occurred is written

P (A|B)

This is read as ‘the conditional probability of A given B’ or ‘the probability of A condi-

tioned on B’ or simply ‘the probability of A given B’.

We can visualize conditional probability as follows. Think of P (A) as the proportion of the

area of the whole sample space taken up by A.For P (A|B) we restrict our attention to B.

That is, P(A|B) is the proportion of area of B taken up by A, i.e. P (A ∩ B)/P (B).

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