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Properties of Outcomes Help (page 2)

By — McGraw-Hill Professional
Updated on Aug 26, 2011

Complementary Outcomes

If two outcomes H1 and H2 are complementary, then the probability, expressed as a ratio, of one outcome is equal to 1 minus the probability, expressed as a ratio, of the other outcome. The following equations hold:

    p(H2) = 1 – p(H1)
    p(H1) = 1 – p(H2)

Expressed as percentages:

    p%(H2) = 100 – p%(H1)
    p%(H1) = 100 – p%(H2)

Figure 3-3 shows this as a Venn diagram.

Properties of Outcomes

The notion of complementary outcomes is useful when we want to find the probability that an outcome will fail to occur. Consider again the election between Mrs. Anderson and Mr. Boyd. Imagine that you are one of those peculiar voters who call themselves ''contrarians,'' and who vote against, rather than for, candidates in elections. You are interested in the probability that ''your candidate'' (the one you dislike more) will lose. According to the pre-election poll, pemp(A) = 0.29 and pemp(B) = 0.71. We might state this inside-out as:

      pempB) = 1 – pemp(B)
            = 1 – 0.71
                = 0.29
      pempA) = 1 – pemp(A)
            = 1 – 0.29
                = 0.71

where the ''droopy minus sign'' (¬) stands for the ''not'' operation, also called logical negation. If you are fervently wishing for Mr. Boyd to lose, then you can guess from the poll that the likelihood of your being happy after the election is equal to pempB), which is 0.29 in this case.

Note that in order for two outcomes to be complementary, the sum of their probabilities must be equal to 1. This means that one or the other (but not both) of the two outcomes must take place; they are the only two possible outcomes in a scenario.

Nondisjoint Outcomes

Outcomes H1 and H2are called nondisjoint if and only if they have at least one element in common:

    H1 H2

In this sort of case, the probability of either outcome is equal to the sum of the probabilities of their occurring separately, minus the probability of their occurring simultaneously. The equation looks like this:

    p(H1 H2) = p(H1) + p(H2) – p(H1 H2)

Figure 3-4 shows this as a Venn diagram. The intersection of probabilities is subtracted in order to ensure that the elements common to both sets (represented by the lightly shaded region where the two sets overlap) are counted only once.

Properties of Outcomes

Multiple Outcomes

The formulas for determining the probabilities of mutually exclusive and nondisjoint outcomes can be extended to situations in which there are three possible outcomes.

Three mutually exclusive outcomes. Let H1, H2, and H3be three mutually exclusive outcomes, such that the following facts hold:

    H1 H2 =
    H1 H3 =
    H2 H3 =

The probability of any one of the three outcomes occurring is equal to the sum of their individual probabilities (Fig. 3-5):

Properties of Outcomes

    p(H1 H2 H3) = p(H1) + p(H2) + p(H3)

Three nondisjoint outcomes. Let H1, H2, and H3 be three nondisjoint outcomes. This means that one or more of the following facts is true:

    H1 H2 =
    H1 H3 =
    H2 H3 =

The probability of any one of the outcomes occurring is equal to the sum of the probabilities of their occurring separately, minus the probabilities of each pair occurring simultaneously, plus the probability of all three occurring simultaneously (Fig. 3-6):

Properties of Outcomes

              p(H1 H2 H3)
          = p(H1) + p(H2) + p(H3)
    p(H1 H2) – p(H1 H3) – p(H2 H3)
          + p(H1 H2 H3)

Properties of Outcomes Practice Problems

Practice 1

Imagine that a certain high school has 1000 students. The new swimming and diving coach, during his first day on the job, is looking for team prospects. Suppose that the following are true:

  • 200 students can swim well enough to make the swimming team
  • 100 students can dive well enough to make the diving team
  • 30 students can make either team or both teams

If the coach wanders through the hallways blindfolded and picks a student at random, determine the probabilities, expressed as ratios, that the coach will pick

  • a fast swimmer; call this p(S)
  • a good diver; call this p(D)
  • someone good at both swimming and diving; call this p(S D)
  • someone good at either swimming or diving, or both; call this p(S D)

Solution 1

This problem is a little tricky. We assume that the coach has objective criteria for evaluating prospective candidates for his teams! That having been said, we must note that the outcomes are not mutually exclusive, nor are they independent. There is overlap, and there is interaction. We can find the first three answers immediately, because we are told the numbers:

      p(S) = 200/1000 = 0.200
      p(D) = 100/1000 = 0.100
    p(S D) = 30/1000 = 0.030

In order to calculate the last answer – the total number of students who can make either team or both teams – we must find p(S D) using this formula:

    p(S D) = p(S) + p(D) – p(S D)
        = 0.200 + 0.100 – 0.030
          = 0.270

This means that 270 of the students in the school are potential candidates for either or both teams. The answer is not 300, as one might at first expect. That would be the case only if there were no students good enough to make both teams. We mustn't count the exceptional students twice. (However well somebody can act like a porpoise, he or she is nevertheless only one person!)

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