Basic properties of event probabilities

Here are some useful properties about event probabilities. We will need them again and again.

Theorem 1

Consider three events EE, FF and GG of a random experiment. SS denotes the sample space of the random experiment.

  1. p({})=0p(\{ \})=0
  2. p(S)=1p(S)=1
  3. p(E∪F)=p(E)+p(F)−p(E∩F)p(E\cup F)=p(E)+p(F)-p(E\cap F)
  4. p(E∪F)=p(E)+p(F)p(E\cup F)=p(E)+p(F) if EE and FF are mutually exclusive.
  5. p(E∪F∪G)=p(E)+p(F)+p(G)p(E\cup F \cup G)=p(E)+p(F)+p(G) if EE, FF and GG are pairwise mutually exclusive. It is straight forward to generalise to an arbitrary number of pairwise mutually events.
  6. p(E′)=1−p(E)p(E^\prime)=1-p(E)
Exercise 1

Give a proof of the statements above.

Solution

  1. The event {}\{\} contains not outcomes, and as a random experiment always produces exactly one outcome, this event never happens. The relative frequency of the occurrence of this event is therefore zero, thus p({})=0p(\{ \})=0.

  2. The event SS contains all possible outcomes, and as a random experiment always produces one outcome, this event occurs every time. So p(S)=1p(S)=1.

  3. The statement follows from the figure below:

  4. As EE and FF are mutually exclusive, if is not possible that EE and also FF occur in the same experiment. Thus p(E∩F)=0p(E \cap F)=0 and therefore, with (3),

    p(E∪F)=p(E)+p(F)−0=p(E)+p(F)p(E\cup F)=p(E)+p(F)-0=p(E)+p(F)

    Note that we can also argue that p(E∩F)=0p(E\cap F)=0 because E∩F={}E\cap F=\{ \}.

  5. The events EE and H=F∪GH=F\cup G are mutually exclusive, and therefore

    p(E∪F∪G)=p(E∪H)=p(E)+p(H)=p(E)+p(F∪G)=p(E)+p(F)+p(G)\begin{array}{lll} p(E\cup F\cup G) &= & p(E\cup H)\\ &=&p(E)+p(H)\\ &=& p(E)+p(F\cup G)\\ &=& p(E)+p(F)+p(G) \end{array}

    Another argument, is as follows: Let us repeat the experiment NN times. Because EE, FF and GG will never occur in the same experiment, the percentage of times that p(E∪F∪G)p(E\cup F\cup G) occurs must be the sum of the percentages p(E)+p(F)+p(G)p(E)+p(F)+p(G).

  6. It is p(E∪E′)=p(S)=1p(E\cup E^\prime)=p(S)=1 and because EE and E′E^\prime are mutually exclusive, we have 1=p(E∪E′)=p(E)+p(E′)1=p(E\cup E^\prime)=p(E)+p(E^\prime), and thus p(E′)=1−p(E)p(E^\prime)=1-p(E).

    We could also proof the statement directly. Let's repeat the experiment NN times (NN large). For every repetition either EE or the opposite event E′E^\prime will occur. Thus p(E)+p(E′)=1p(E)+p(E^\prime)=1, and therefore p(E′)=1−p(E)p(E^\prime)=1-p(E).

Note that the above statements become obvious if we represent the events in a Venn-diagram, and identify the circle areas with the probability of an event to happen. The rectangle indicating SS has area 11.

Using Venn-diagrams in this way helps to solve problems as well. See the exercise below.

Exercise 2

Consider a random experiment where event AA has probability 0.60.6 to occur, and event BB has probability 0.30.3 to occur. The probability that AA and BB occur in the same experiment is 0.20.2. Determine the probability that

  1. AA or BB occur.

  2. BB does not occur.

  3. AA occurs and BB does not occur.

  4. neither AA nor BB occur.

Solution

Draw the Venn-diagram and indicate the probabilities, starting with the intersection p(A∩B)=0.2p(A\cap B)=0.2.

  1. p(A∪B)=0.4+0.2+0.1=0.7‾p(A\cup B)=0.4+0.2+0.1=\underline{0.7}
  2. p(B′)=1−0.3=0.7‾p(B^\prime)=1-0.3=\underline{0.7}
  3. p(A∩B′)=0.4‾p(A\cap B^\prime)=\underline{0.4}
  4. p((A∪B)′)=1−p(A∪B)=1−0.7=0.3‾p((A\cup B)^\prime) =1-p(A\cup B)=1-0.7=\underline{0.3}