Satyagopal Mandal
Department of Mathematics
University of Kansas
Office: 624 Snow Hall  Phone: 785-864-5180
  • e-mail: mandal@math.ukans.edu
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    Chapter 15: Probability 2

     

    Now we are ready to talk about probability of an outcome or an event. If we toss a coin, one believes that the probability that the face H will show up is 1/2. But this is about a "normal" coin. What if we toss a loaded coin? If you have a loaded coin, you may know that the probability that the face H will show up is 1/5. What does all these mean? How and where did you learn that, for your loaded die, the probability that the face H will show up is 1/5? When an ordinary person makes such probability statements he/she is, in fact, talking about his/her experience. Regarding your loaded coin, you have tossed your coin many-many times and have experienced that about once in 5 tosses the face H showed up and other times the face Tail showed up. So, "you know" that the probability that the face H will show up is 1/5.

     

    Similarly, you rolled a die many-many times. You have experienced that about once in 6 throws the face 4 shows up. So, "you know" that probability that the face 4 will show up when you throw the die is 1/6. It will be a different story if you are working with a loaded die, because your experience would tell you something different.

     

    That was the "real life" idea of probability. In the study of the mathematics of probability, we assume this "experience part" and do the mathematics.

     

     

     

    The mathematics of probability includes the following:

    1. Description of the sample space S (and/or the random experiment).

    2. A method or formula to compute the probability of an event. So, given an event E, the method or the formula will give us how to compute the probability P(E) of E. The probabilities P(E) must satisfy certain laws of probability.

     

    A probability space is a sample S space with probability assignment as in 2). So, to describe a probability space, we have to give 1) and 2).

     

    Laws of Probability: Let

    S = {o1, o2, …, oN}

    be a (finite) sample space. (Here o1, o2, …, oN are the outcomes of the experiments.) Following are the elements of probability spaces:

    1. For each outcome oi, a method or a formula is given so that we can compute the number P(oi), to be called the probability that the outcome was oi. The probabilities P(oi) must satisfy the following two properties:

    1. P(oi) is a number between 0 and 1;

    2. P(o1)+ P(o2)+…+ P(oN),=1.

    So, the sum of probabilities of all the outcomes is 1.

     

    2) Given an event E, the probability P(E) of E is defined to be the sum of probabilities of all the outcomes in E.

    So,

    P(E) = the sum of probabilities of all the outcomes in E.

    1. The probability of the impossible event is zero and the probability of the certain event is 1.

    P(impossible event) = 0 and

    P(certain event) = P(S) = 1.

     

    Following are some examples of probability spaces.

     

    Example 1: Suppose we roll a (loaded) die. So, the sample space is S={1, 2, 3, 4, 5, 6}. We are given that

    P(1) = P(2) = P(3) = P(4) = 1/12 and P(5) = P(6) = 1/3.

     

    Find the probability of the event E={1, 3, 5}. By 2) we have P(E) = P(1) + P(3) + P(5) = 1/12 + 1/12 + 1/3 = 1/2.

     

    Find the probability of the event F={1,6}.

    P(F) = P(1) + P(6) = 1/12 + 1/3 = 5/12.

     

    Remark. Please note that here all the outcomes did not have the same probability.

     

    Probability Spaces with Equally Likely Outcomes:

     

    Unlike the above example, there are probability spaces where every outcome has equal probability. (Such is the case when you toss a "normal" coin or roll a "normal" die.) In such cases, we say that outcomes are equally likely.

     

    When outcomes are equally likely, the probability P(E) can be computed by counting the number of outcomes in E and that of S.

     

    If S has N outcomes then we have:

     

    1. P(individual outcome) = P(oi) = 1/N

    2. For an event E

    P(E) = (# of outcomes in E)/(# of outcomes in S)

    =(# of outcomes in E)/N

     

    Remark: Outcomes are equally likely in "normal" situations. Examples are unbiased coins ("honest coins", as you textbook calls them), fair dice, picking a card from a shuffled deck of cards and so on.

     

     

    Suggested Problems: Examples 18-26 (pp. 496-), Ex 1d, 2c, 2e, 13a-b, 14c, 15a-b, 16a; 22.

     

    Example 2: (Compare with Ex. 13). Suppose we roll a (fair) die 3 times.

    1. Describe the sample space.

    2. Count the number of outcomes in the sample space S.

    3. What is the probability that sum of the points on the 3 faces is 7?

    4. What is the probability that of the sum of the points on the faces is not equal to 7?

     

    1) Answer is S={(1,1,1), (1,1,2), (1,1,3), (1,1,4), (1,1,5), (1,2,6), (1,2,1), (1,2,2), (1,2,3) ….}.

     

    1. The job of rolling a die 3 times can be done in 3 stages.

      2.  

       

      Stage

      The job

      Number of outcomes

      1

      Roll the die

      6

      2

      Roll the die again

      6

      3

      Roll the die again

      6

       

      So, by multiplication rule, the number of outcomes in S is 63.

       

    2. Let E be the event that the sum is 7. So,

      3.  

      E = {(1,1,5), (1,2,4), (1,3,3), (1,4,2), (1,5,1),

      (2,1,4), (2,2,3), (2,3,2), (2,4,1),

      (3,1,3), (3,2,2), (3,3,1),

      (4,1,2), (4,2,1),

      (5,1,1)}

      So E has 15 outcomes. Therefore,

      P(E) = (# of outcomes in E)/(# of outcomes in S)

      = 15/63 = 15/216.

       

    3. Let F be the event that the sum is not equal to 7. So, the

     

    # of outcomes in F

    = (# of outcomes in S) -( # of outcomes in E)

    = 216 - 15 = 201.

    Therefore

     

    P(F) = (# of outcomes in F)/(# of outcomes in S) =201/216.

     

     

    Example 3: (Compare with Ex. 22.) Suppose we have to form a committee of 2 from of group of 15 men and 19 women.

    1. Describe the sample space and count the number of outcomes in S.

    2. What is the probability that both the members of the committee are men?

    3. What is the probability that both the members are women?

    4. What is the probability that one member is man and the other member is woman?

     

    This is a problem of unordered-selection.

    1. S is the set of all possible pair selected from this group of 15+19 = 34 people. So the number of outcomes in S =34C2= 34P2/2! =(34 x 33)/1x2 = 561.

    2. Let E be the event that both the members of the committee are men. So,

      3.  

      the number of outcomes in E = number of ways we can pick 2 from a group of 15 men = 15C2= 15P2/2! = (15x14)/2 = 105. Therefore

       

      P(E) = (# of outcomes in E)/(# of outcomes in S) = 105/561.

       

    3. Let F be the event that both members of the committee are women. So,

      4.  

      the number of outcomes in F = number of ways we can pick 2 from a group of 19 women = 19C2= 19P2/2! = (19x18)/2 = 171. Therefore

       

      P(F) = (# of outcomes in F)/(# of outcomes in S) =171/561.

    4. Let M be the event that one member is man and the other is a woman. The job of picking such a committee can be done in two stages:

     

    Stage

    The job

    Number of ways

    1

    Pick a male member

    15

    2

    Pick a female member

    19

     

    So, the number of outcomes in M = 15x19 = 285.

    So,

    P(M) = (# of outcomes in M)/(# of outcomes in S)

    = 285/561.