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 3

     

    The Roundabout Approach

     

    Sometime it is easier to find the probability that an event E does not occur. Then, we can use this to compute the probability that E occur. This topic (or the formula to come) was not discussed explicitly in your textbook.

     

    Definition: Suppose S is the sample space and E is an event. Then (not E) will denote the event that E does not occur. (not E) is also called complement of E or the opposite event of E.

     

    Formula: We have

    P(E) + P(not E) = 1

    Or

    P(E) = 1 - P(not E).

     

    Example 1.1: (Compare with Ex. 19 a, pp.506.) Suppose we roll a die 3 times. What is the probability that the face 6 will show up at least once?

     

    Let E be the event that the face 6 will show up at least once. We know that the sample space S has 63 outcomes. But it will be easier to count the number of outcomes in (not E). We have (not E) is the event that the face 6 never showed up in the 3 rolls. This can happen in the following 3 stages:

     

    Stage

    The job

    Number of ways

    1

    Roll 1- 5 in 1st roll

    5

    2

    Roll 1-5 in 2nd roll

    5

    3

    Roll 1-5 in 3rd roll

    5

     

    So the number of outcomes in (not E) = 53.

    So, the probability

    P(not E) = (# of outcomes in (not E))/(# of outcomes in S) = 53/63=125/216.

    So,

    P(E) = 1 - P(not E) = 1 - 125/216 = 91/216.

     

    Example 1.2: (Compare with Ex. 19 b, pp.506.) Suppose we roll a die 7 times. What is the probability that an even-number face will show up at least once?

     

    The experiment can be performed by rolling the die 7 times, (or in "7 stages"). So, by multiplication rule the sample space S has 67 outcomes.

    Just remember that opposite of "at least once" is "never". Let us denote by E the event that an even-number face will show up at least once. So, (not E) is the event that an even-number face never showed up. An outcome in (not E) can be accomplished in the following seven stages:

     

    Stage

    The job

    Number of ways

    1

    Roll 1,3,5 in 1st roll

    3

    2

    Roll 1,3,5 in 2nd roll

    3

    3

    Roll 1,3,5 in 3rd roll

    3
     

    Roll 1,3,5 in 4th roll

    3
     

    Roll 1,3,5 in 5th roll

    3
     

    Roll 1,3,5 in 6th roll

    3
     

    Roll 1,3,5 in 7th roll

    3

    So the number of outcomes in (not E) = 37.

    So, the probability

    P(not E) = (# of outcomes in (not E))/(# of outcomes in S) = 37/67.

     

    So,

    P(E) = 1 - P(not E) = 1 -37/67.

     

    Suggested Problems: Ex. 19 (pp. 506).

     

     

    Independent Events:

     

    Sometime it is understandable that two events E and F do not influence the occurrence of each other. For example, if you roll a die twice and E is the event that first roll will show an odd-number face and F is the event that the second roll will show 1 or 2. Then the occurrence of E will not influence the occurrence of F. (Describe E and F in brace notation.)

     

    Definition: We say that the two events E and F are mutually independent if the occurrence one does not influence the occurrence of the other.

     

    The Multiplication Principle: Suppose E and F are two independent events. Then the probability that both E and F occur is the product P(E)P(F).

    So, if E and F are independent then

    P(E and F) = P(E)P(F).

    Now we are going to use this multiplication principle to compute some probabilities.

     

    Example 2.1: Suppose you are dealt with a hand of 5 cards out of a shuffled deck of 20 high-cards. (Ace, King, Queen and Jack and 10 are the high-cards.)

    1. What is the probability that you will receive all the four aces?

    2. Suppose you are dealt such a hand of 5 cards twice. What is the probability that you will receive all the four aces in both the deals?

     

    1. Here the random experiment is to deal a hand of 5 cards out of 20 high-cards. So, the sample space is the set of all possible combination of 20 cards taken 5 at a time. So, the number of outcomes in S is

    20C5 =20P5/5!=

    (20 x 19 x 18 x 17 x 16)/(1 x 2 x 3 x 4 x 5) = 15504.

    Now let E be the event that a hand of 5 cards has all four aces. Such a hand could be dealt in two stages:

     

    Stage

    The job

    Number of ways

    1

    Pick 4 aces from 4

    4C4=1

    2

    Pick 1 from the remaining 16

    16C1

     

    So, number of outcomes in E = 1 x 16C1= 16C1=16.

    So, P(E) = (# of outcomes in E)/(# of outcomes in S) = 16/15504.

     

    1. Now our experiment is to deal a hand of 5 cards out of 20 cards twice. Let W be the event that we get all four aces twice. Let E1 be the event that we get all four aces in the first deal and E2 be the event that we get all four aces in the second deal. Then

    W = (E1and E2).

    Also from a) we have

    P(E1) = P(E2) = 16/15504.

    Also E1, E2 are mutually independent. So, by multiplication rule we have

    P(W) = P(E1)P(E2) = (16/15504)2.

     

    Example 2.2: Suppose you roll 4 fair dice. What is the probability that at least one die will show the face 3?

     

    Suppose E is the event that at least one die show the face 3. Here the sample space S has 64 outcomes. As I said before, opposite of "at least once" is "never". So,

    (not E) is the event that none of the 4 dice will show the face 3.

     

    Suppose F1 is the event that the 1st die will not show the face 3.

    Suppose F2 is the event that the 2nd die will not show the face 3.

    Suppose F3 is the event that the 3rd die will not show the face 3.

    Suppose F4 is the event that the 4th die will not show the face 3.

     

    We can see that F1, F2, F3, F4 are mutually independent and

    (not E) = (F1 and F2 and F3 and F4).

    We also have

    P(F1) = P(F2) = P(F3) = P(F4 ) = 5/6.

    So,

    P(not E) = P(F1) x P(F2) x P(F3) x P(F4 ) = (5/6)4.

    So,

    P(E) = 1 - P(not E) = 1 - (5/6)4.

     

    Odds for and against:

     

    In many situations probability of an event E is described as "odds" in favor or against. This language is often used in gambling.

     

    Definition 1: Suppose E is an event. We say that odds in favor of E is "m to n" to mean that

    P(E) = m/(m + n).

     

    So, if P(E) = a/b then odds in favor of E is "a to b-a".

     

    Definition 2: We say odds against the event E is "n to m", if odds in favor of E is "m to n".

     

    Suggested Problems: Example 23-29 (pp. 500-); Ex. 13c-d, 14d, 15d, 16a-b, 17a-c, 18a-c, 20.

     

    Example 3.1: (see Ex. 16.) Suppose you roll a die twice.

    1. What is the probability of the event E that "at least one of the two rolls will show 4"?

    2. What are the odds in favor of the event E?

     

    1. First we will compute P(not E). But (not E) is the event that the face 4 did not show up in these two rolls.

      2.  

      Let F1 be the event that face 4 did not show up in the 1st roll.

      Let F2 be the event that face 4 did not show up in the 2nd roll.

       

      To compute P(F1) we just have to look at the first roll. So,

      P(F1) = (# of outcomes in 1st roll that is not 4)/(# of outcomes in the first roll) = 5/6.

      Similarly, P(F2) = 5/6.

      Also (not E) = (F1 and F2)

      So, P(not E) = P(F1 and F2) = P(F1) P(F2) = (5/6)(5/6) = 25/36.

      So, P(E) = 1 - P(not E) = 1 -25/36 = 11/36.

       

    2. Since P(E) = 11/36, we have odds in favor of E is 11 to (36-11).

     

    So, the odds in favor of the event E is 11 to 25.

     

    Example 3.2: (see Ex. 18.) Suppose we have a game where we roll two dice. We win if the sum of the "face values" is less or equal to 5, otherwise we lose.

    1. Find the probability of winning. Also what are the odds in favor of winning?

    2. If we roll the dice twice (i. e. we play twice), what is the probability that we win in both the rolls?

    3. If we roll the dice twice (i. e. we play twice), what is the probability that we lose in both the rolls?

     

    1. Here the experiment is to roll two dice. So, the sample space has 36 outcomes. Let E be the event that you win. Then

      2. E= {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}. So, E has 10 outcomes in it. So,

      P(E)=(# of outcomes in E)/(# of outcomes in S) = 10/36.

      So, the odds in favor of winning are 10 to 26 (=36-10).

       

    2. Let E1 be the event that you win in the first rolls and E2 be the event that you win in the 2nd rolls. Let W be the event that you win both the rolls. Then

      3. W = (E1 and E2)

      And

      P(W) = P(E1 and E2) = P(E1) P(E2) = (10/36)(10/36) = 100/1296.

    3. Let L1 be the event that you lose in the first rolls and L2 be the event that you lose in the 2nd rolls. Let L be the event that you lose both the rolls. Then

    L = (L1 and L2)

    And

    P(L) = P(L1 and L2) = P(L1) P(L2)

    We also have

    P(L1) = 1 - P(E1) = 1 -10/36 = 26/36

    and similarly

    P(L2) = 26/36.

    So,

    P(L) = P(L1) P(L2) = (26/36)(26/36)

    Some Extra Problems:

     

    Example 4.1: A person owns two stocks. The probability of the event that at least one stock will go up in price on a particular day is P(E) = 0.9. What is the probability that none will go up in price on a particular day?

     

    Answer: P(not E) = ?

     

    Example 4.2:Two university employees (Mr. Park and Mr. Jones) issue tickets to illegally parked cars. Probability of the event E that Mr. Jones will notice an illegally parked car is P(E)= 0.1 and the probability of the event F that Mr. Park will notice an illegally parked car is P(F) = 0.3.

    1. What is the probability P(not E) that Mr. Jones will miss an illegally parked car?

    2. What is the probability P(not F) that Mr. Park will miss an illegally parked car?

    3. Assuming independence, what is the probability that both will miss an illegally parked car?

    4. What is the probability that at least one of them will notice an illegally parked car?