MA121: Introduction to Statistics

LEARNING OBJECTIVES

1. To learn how some events are naturally expressible in terms of other events.
2. To learn how to use special formulas for the probability of an event that is expressed in terms of one or more other events.

Some events can be naturally expressed in terms of other, sometimes simpler, events.

Complements

Definition

The complement of an event $A$ in a sample space $S$, denoted $A^c$, is the collection of all outcomes in $S$ that are not elements of the set $A$. It corresponds to negating any description in words of the event $A$.

EXAMPLE 10

Two events connected with the experiment of rolling a single die are $E$ : "the number rolled is even" and $T$ : "the number rolled is greater than two". Find the complement of each.

Solution:

In the sample space $S=\{1,2,3,4,5,6\}$ the corresponding sets of outcomes are $E=\{2,4,6\}$ and $T=\{3,4,5,6\}$. The complements are $E^{c}=\{1,3,5\}$ and $T^{c}=\{1,2\}$.

In words the complements are described by "the number rolled is not even" and "the number rolled is not greater than two". Of course easier descriptions would be "the number rolled is odd" and "the number rolled is less than three".

If there is a 60% chance of rain tomorrow, what is the probability of fair weather? The obvious answer, 40%, is an instance of the following general rule.

Probability Rule for Complements

$P\left(A^{c}\right)=1-P(A)$

This formula is particularly useful when finding the probability of an event directly is difficult.

EXAMPLE 11

Find the probability that at least one heads will appear in five tosses of a fair coin.

Solution:

Identify outcomes by lists of five $h s$ and $t$, such as $t t h t t$ and $h h t t t$. Although it is tedious to list them all, it is not difficult to count them. Think of using a tree diagram to do so. There are two choices for the first toss. For each of these there are two choices for the second toss, hence $2 \times 2=4$ outcomes for two tosses. For each of these four outcomes, there are two possibilities for the third toss, hence $4 \times 2=8$ outcomes for three tosses. Similarly, there are $8 \times 2=16$ outcomes for four tosses and finally $16 \times 2=32$ outcomes for five tosses.

Let $O$ denote the event "at least one heads". There are many ways to obtain at least one heads, but only one way to fail to do so: all tails. Thus although it is difficult to list all the outcomes that form 0 , it is easy to write $O^{c}=\{t t t t t\}$. Since there are 32 equally likely outcomes, each has probability $1 / 32$, so $P\left(O^{c}\right)=1 / 32$, hence $P(O)=1-1 / 32 \approx 0.97$ or about a $97 \%$ chance.