Random events

Consider a random experiment. The set SS of all possible outcomes is called the sample space of the experiment. Thus, if the experiment has mm possible outcomes denoted by o1,...,omo_1, ..., o_m, then the sample space is

S={o1,...,om}S=\{ o_1, ..., o_m\}

Any subset ESE\subset S is called an event. Let's perform the experiment. We say that event E has occurred if the outcome that occurred is in EE.

Some special events

Example 1
  1. A coin is flipped. The sample space is

    S={H,T}S=\{H,T\}
  2. A die is rolled. The sample space is

    S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\}
    • The "event E={2,4,6}E=\{2,4,6\} occurs" can also expressed as "an even number occurs".
    • The opposite event is E={1,3,5}E^\prime=\{1,3,5\}, that is, the event that "an odd number occurs".
    • That "a number smaller than 33 was observed" can also be expressed as "the event E={1,2}E=\{1,2\} occurred".
  3. A coin is flipped twice. The sample space is

    S={HH,HT,TH,TT}S=\{HH,HT, TH, TT\}

    where we mean with HHHH that HH occurred in the first flip, and HH in the second flip. And so on.

    • The observation that "exactly one head occurred" can also be expressed as "the event E={HT,TH}E=\{HT, TH\} occurred".
    • The event "no head occurred" can also be expressed as "the event E={TT}E=\{TT\} occurred".
    • The opposite event is E={HT,TH,HH}E^\prime = \{HT, TH, HH\}, that is, the event that at least one head occurs.
  4. Random selection of a person in a group of mm people. The sample space consists of all possible people in the group:

    S={p1,p2,...,pm}S=\{p_1, p_2, ... , p_m\}

    An event could be EE="person has green eyes". This event EE contains then all persons with green eyes.

Exercise 1

Determine the sample space and the number of outcomes for the following random experiments:

  1. Flipping a coin three times.

  2. Rolling a die twice.

Solution
  1. S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT},S=8S=\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}, \vert S \vert = 8.
  2. S={11,12,13,14,15,16,...,61,62,63,64,65,66},S=36S=\{11,12,13,14,15,16, ..., 61, 62, 63, 64, 65, 66\}, \vert S \vert = 36. Note that with "1515" we mean a "11" in the first roll and a "55" in the second roll, and so on.
Exercise 2

Express each event as a selection of outcomes.

  1. You roll a die twice.

    EE="the sum of the two observed numbers is even"

    FF="at least one 6 occurred"

    GG="no six occurred"

    HH="the sum is between 66 and 88 (including 66 and 88)"

  2. You randomly select a number between 11 and 1010 from a box

    II="the selected number is a prime"

    JJ="the selected number is divisible by 33 and bigger than 55"

Solution

E={11,13,15,22,24,26,31,33,35,42,44,46,51,53,55,62,64,66}E=\{11,13,15,22,24,26,31,33,35,42,44,46,51,53,55,62,64,66\}

F={16,26,36,46,56,61,62,63,64,65,66}F=\{16,26,36,46,56,61,62,63,64,65,66\}

G=FG=F^\prime

H={15,16,24,25,26,33,34,35,42,43,44,51,52,53,61,62}H=\{15,16,24,25,26,33,34,35,42,43,44,51,52,53,61,62\}

I={2,3,5,7}I=\{2,3,5,7\}

J={6,9}J=\{6,9\}