iii If A 1 A n are mutually exclusive events then P A 1 A n P A 1 P A n A set

Iii if a 1 a n are mutually exclusive events then p a

This preview shows page 89 - 97 out of 564 pages.

(iii). If A 1 , . . . , A n are mutually exclusive events, then P ( A 1 . . . A n ) = P ( A 1 ) + . . . + P ( A n ) . A set of events are mutually exclusive if at most one of the events can occur in a given experiment. This axiom says that to calculate the probability of the union of distinct events we can simply add their individual probabilities. 2.3.2 Complement Law If A is an event, the set of all outcomes that are not in A is called the complement of the event A, denoted A { . This is (pronounced ‘A complement’).
Image of page 89
Probability I 79 The rule is A { = 1 - P ( A ) (Law 1) Example 2.4: Complements Let S (the sample space) be the set of students at Oxford. We are picking a student at random. Let A = The event that the randomly selected student suffers from depression We are told that 8% of students suffer from depres- sion, so P ( A ) = 0 . 08 . What is the probability that a student does not suffer from depression? The event { student does not suffer from depression } is A { . If P ( A ) = 0 . 08 then P ( A { ) = 1 - 0 . 08 = 0 . 92 .
Image of page 90
Probability I 80 2.3.3 Addition Law (Union) Suppose, A = The event that a randomly selected student from the class has b B = The event that a randomly selected student from the class has b What is the probability that a student has brown eyes OR blue eyes? This is the union of the two events A and B, denoted A B (pronounced ‘A or B’) We want to calculate P(A B). In general for two events P(A B) = P(A) + P(B) - P(A B) (Addition Law) To understand this law consider a Venn diagram of the situa- tion (below) in which we have two events A and B . The event A B is represented in such a diagram by the combined sam-
Image of page 91
Probability I 81 ple points enclosed by A or B . If we simply add P ( A ) and P ( B ) we will count the sample points in the intersection A B twice and thus we need to subtract P ( A B ) from P ( A )+ P ( B ) to calculate P ( A B ) . A B A B S Example 2.5: SNPs Single nucleotide polymorphisms (SNPs) are nu- cleotide positions in a genome which exhibit varia- tion amongst individuals in a species. In some stud- ies in humans, SNPs are discovered in European populations. Suppose that of such SNPs, 70% also
Image of page 92
Probability I 82 show variation in an African population, 80% show variation in an Asian population and 60% show vari- ation in both the African and Asian population. Suppose one such SNP is chosen at random, what is the probability that it is variable in either the African or the Asian population? Write A for the event that the SNP is variable in Africa, and B for the event that it is variable in Asia. We are told P ( A ) = 0 . 7 P ( B ) = 0 . 8 P ( A B ) = 0 . 6 .
Image of page 93
Probability I 83 We require P ( A B ) . From the addition rule: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 7 + 0 . 8 - 0 . 6 = 0 . 9 .
Image of page 94
3 Probability II 3.1 Independence and the Multiplication Law If the probability that one event A occurs doesn’t affect the probability that the event B also occurs, then we say that A and B are independent . For example, it seems clear than one coin doesn’t know what happened to the other one (and if it did know, it wouldn’t care), so if A 1 is the event that the first coin comes up heads, and A 2 the event that the second coin comes up heads, then 84
Image of page 95
Probability II 85 Example 3.1: One die, continued
Image of page 96
Image of page 97

You've reached the end of your free preview.

Want to read all 564 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture