2 Definitions of probability

2.1 Relative frequency applied to probability

Consider the experiment of throwing a single coin many times.

Suppose we throw a coin 10 times and obtain six Heads and four Tails; does this suggest that the coin is biased? Clearly not! What about the case when we obtain 9 Heads and 1 Tail?

We conducted an experiment in which a coin was thrown 100 times and the result recorded each time as 1 if a Head appeared face up and 0 if a Tail appeared. In Figure 7 we have plotted the average score r n , where r is the number of Heads and n is the total number of throws, against n for n = 10 , 20 , … , 100 . The quantity r n is called the relative frequency of Heads.

Figure 7

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As n increases the relative frequency settles down near the value 1 2 . This is an experimental estimate of the probability of throwing a Head with this particular coin. Note that when n = 50 this estimate was 0.49 and when n = 100 this estimate was 0.51. When we repeated the whole experiment again, the value of r n when n = 100 was 0.46. Hence the use of the word estimate . Normally, as the number of trials is increased the estimate tends to settle down but this is not certain to occur.

Theoretically, the probability of obtaining a Head when a fair coin is thrown is 1 2 . Experimentally, we expect the relative frequency to approach 1 2 as n increases.

2.2 Equi-probable spaces and the principle of equally likely outcomes

An equi-probable space is a sample space in which the chance that any one sample point occurs is equal to the chance that any other sample point occurs. Whether a sample space is an equi-probable space is usually determined by inspection or logic.

  1. Tossing a coin: the sample space S is, using an obvious notation:

    S = { H , T }

    H and T are the two simple outcomes and are equally likely to occur.

  2. Rolling a die: a sample space is:

    S = { 1 , 2 , 3 , 4 , 5 , 6 }

    Each number from 1 to 6 is a simple outcome and is equally likely to occur.

  3. Tossing three coins: a sample space comprised of simple outcomes is:

    S = { H H H , H H T , H T H , H T T , T H H , T H T , T T H , T T T }

    Each of the eight outcomes stated is equally likely to occur.

  4. Counting the number of heads when three coins are tossed. Here S = { 3 Heads , 2 Heads , 1 Head , 0 Heads } . This is not an equi-probable space and the outcomes are not equally likely. (For example: the event {2 Heads} is the union of three simple events { H H T , H T H , T H H } so must occur more often than the event { H H H } .

Of the sample spaces above all are equi-probable spaces except for (4).

Key Point 3

The Principle of Equally Likely Outcomes

This states that each simple outcome in an equi-probable space is equally likely to occur. This principle enables us to deduce the probabilities that simple events (and hence more complicated events which are combinations of simple events) occur.

Notation

If A is an event associated with a sample space S the the probability of A occurring is denoted by P ( A ) .

Referring to the examples above we may immediately deduce that

  1. P { H } = P { T } = 1 2
  2. P { 1 } = P { 2 } = P { 3 } = P { 4 } = P { 5 } = P { 6 } = 1 6
  3. P { H H H } = P { H H T } = P { H T H } = P { T H H } =

    P { H T T } = P { T H T } = P { T T H } = P { T T T } = 1 8

2.3 Definition

We can now define probability using the Principle of Equally Likely Outcomes as follows:

If a sample space S consists of n simple outcomes which are equally likely and an event A consists of m of those simple outcomes, then

P ( A ) = m n = the number of simple outcomes in  A the number of simple outcomes in  S

It follows from this definition that 0 ≤ P ( A ) ≤ 1.

The set with no outcomes in it is called the empty set and written ∅ ; therefore P ( ∅ ) = 0 .

Task!

For each of the following events A , B , C , list and count the number of outcomes it contains and hence calculate the probability of A , B or C occurring.

    1. A = “throwing 3 or higher with one die”,
    2. B = “throwing exactly two Heads with three coins”,
    3. C = “throwing a total score of 14 with two dice”.
    1. There are six possible equally likely outcomes of the experiment and four of them, { 3 , 4 , 5 , 6 } , constitute the event A ; hence P ( A ) = 4 6 = 2 3 .
    2. There are eight equally likely outcomes of which three, { H H T , H T H , T H H } are elements of B ; hence P ( B ) = 3 8 .
    3. It is impossible to throw a total higher than 12 so that C = ∅ and P ( C ) = 0 .

Not surprisingly, the probabilities of an event A and its complement are related. The probability of the event A ′ is easily found from the identity

number of outcomes in  A total number of outcomes + number of outcomes not in  A total number of outcomes ≡ 1

so that P ( A ) + P ( A ′ ) ≡ 1

Key Point 4

The Complement Rule

P ( A ′ ) = 1 − P ( A )
In words:

The probability of the complement of A occurring

is equal to 1 minus the probability of A occurring.

Task!

For the events in (a) and (b) of the previous Task find P ( A ′ ) and P ( B ′ ) and describe in words what A ′ and B ′ are in this case.

    1. P ( A ) = 2 3 so that P ( A ′ ) = 1 3 .

      A ′ is the event of throwing a score of less than 3 on one die.

    2. P ( B ) = 3 8 so that P ( B ′ ) = 5 8

      B ′ is the event of throwing no Heads, exactly one Head or exactly three Heads with three coins.

The use of the event A ′ can sometimes simplify the calculation of the probability P ( A ) . For example, suppose that two dice are thrown and we require the probability of the event

A : that we obtain a total score of at least four.

There are many combinations that produce a total score of at least four; however there are only 3 combinations that produce a total score of two or three which is the complementary event to the one of interest. The event A ′ = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) } (where we use an obvious notation of stating the total score on the first die followed by the score on the second die) is the complement of A .

Now P ( A ′ ) = 3 36 since there are 6 × 6 possible combinations in throwing two dice. Thus

P ( A ) = 1 − P ( A ′ ) = 1 − 3 36 = 33 36 = 11 12 .

Task!

Find the probability of obtaining a total score of at least five when three dice are thrown. Hint: identify A and A ′ , then calculate P ( A ′ ) , then P ( A ) .

There are 6 × 6 × 6 = 216 possible outcomes. If, for example, ( 1 , 3 , 6 ) denotes the scores of 1 on die one, 3 on die two and 6 on die three and if A is the event ‘a total score of five or more’ then A ′ is the event ‘a total score of less than 5’ i.e.

A ′ = { ( 1 , 1 , 1 ) , ( 2 , 1 , 1 ) , ( 1 , 2 , 1 ) , ( 1 , 1 , 2 ) }

There are four outcomes in A ′ and hence P ( A ′ ) = 4 216 so that P ( A ) = 212 216 .

Exercises
  1. For each of the following experiments, state whether the variable is discrete or continuous. In each case state the sample space.
    1. The number of defective items in a batch of twenty is noted.
    2. The weight, in kg, of lubricating oil drained from a machine is determined using a spring balance.
    3. The natural logarithm of the weight, in kg, according to a spring balance, of lubricating oil drained from a machine, is noted.
  2. An experiment consists of throwing two four-faced dice (regular tetrahedra) with faces labelled 1, 2, 3, 4.
    1. Write down the sample space of this experiment.
    2. If A is the event ‘total score is at least 4’ list the outcomes belonging to A ′ .
    3. If each die is fair find the probability that the total score is at least 6 when the two dice are thrown. What is the probability that the total score is less than 6?
    4. What is the probability that a double: i.e. ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 4 ) will not be thrown?
    5. What is the probability that a double is not thrown and the score is less than 6?
  3. A lot consists of 10 good articles, 4 articles with minor defects and 2 with major defects. One article is chosen at random from the lot. Find the probability that:
    1. it has no defects,
    2. it has no major defects,
    3. it is either good or has major defects.
  4. Propeller shafts for marine applications are inspected to ensure that they satisfy both diameter requirements and surface finish requirements. The results of 400 inspections are as follows:

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    1. What is the probability that a shaft selected at random satisfies the surface finish requirements?
    2. What is the probability that a shaft selected at random satisfies both diameter and surface finish requirements?
    3. What is the probability that a shaft selected at random satisfies either the diameter or the surface finish requirements?
    4. What is the probability that a shaft selected at random satisfies neither the diameter nor the surface finish requirements?
    1. The variable is discrete. The sample space is { 1 , 2 , … , 20 } .
    2. The variable is continuous. The sample space is the set of real numbers x such that 0 ≤ x < ∞ .
    3. The variable is continuous. The sample space is the set of real numbers x such that − ∞ < x < ∞ .
    1. S = { ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 3 , 1 ) , ( 3 , 2 ) , ( 3 , 3 ) , ( 3 , 4 ) , ( 4 , 1 ) , ( 4 , 2 ) , ( 4 , 3 ) , ( 4 , 4 ) }
    2. A ′ = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) }
    3. The outcomes in the event are { ( 2 , 4 ) , ( 3 , 3 ) , ( 3 , 4 ) , ( 4 , 2 ) , ( 4 , 3 ) , ( 4 , 4 ) } so the probability of this event occurring is 6 16 = 3 8 . The probability of the complement event is 1 − 3 8 = 5 8 .
    4. The probability of a double occurring is 4 16 so the probability of the complement (i.e, double not thrown) is 1 − 4 16 = 3 4 .
    5. Here, consider the sample space in (a). If the doubles and those outcomes with a score greater than 6 are removed we have left the event :

      { ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 1 ) , ( 2 , 3 ) , ( 3 , 1 ) , ( 3 , 2 ) , ( 4 , 1 ) } .

      Hence the probability of this event occurring is 8 16 = 1 2 .

  3. Let G be the event ‘article is good’ , M n be the event ‘article has minor defect’ and M j be the event ‘article has major defect’ 
    1. Here we require P ( G ) . Obviously P ( G ) = 10 16 = 5 8
    2. We require P ( M j ′ ) = 1 − P ( M j ) = 1 − 2 16 = 7 8
    3. The event we require is the complement of the event M n .

      Since P ( M n ) = 4 16 = 1 4 we have P ( M n ′ ) = P ( G  or  M j ) = 1 − 1 4 = 3 4 .

      Equivalently P ( G ) + P ( M n ) = 10 16 + 2 16 = 12 16 = 3 4

    1. 250 400 = 0.625
    2. 200 400 = 0.5
    3. 330 400 = 0.825
    4. 70 400 = 0.175