With or Without Replacement

Aims: By the end of this chapter, you will be able to
(i) define exclusive and non-exclusive events,
(ii) define independent and non-independent events,
(iii) employ formula related to the concepts above, and
(iv) solve problems with Venn Diagram.

  1. Two events A and B are exclusive events if they cannot happen at the same time.
    Events A and B are also called disjoint because in a Venn Diagram, set A and set B do not share any intersection.
    That is, if A and B are exclusive then P(A ∩ B)=0 since the intersection between them is empty.
    For example the event of picking a male student from a class to take a class photo and the event of picking a female student to do the same task are exclusive.
    1. In this case, P( A or B ) = P(A) + P(B).
      P(A or B) is often written as P(A ∪ B).
      Thus, P(A ∪ B) = P(A) + P(B).
    2. Example 1:
      The cafetaria has 60 chocolate, 55 vanila, and 35 mango flavoured icecream. Ice-cream of different flavour is given randomly to students. Find the probability that Jonathan receives
      1. a chocolate flavoured icecream.
      2. either a chocolate or vanila flavoured icecream.
      3. either a mango or vanila flavoured icecream.
      Solutions:
      1. P(chocholate) = 60/150 = 2/5.
      2. P (chocolate or vanila) = P(choclate) + P(vanila)
        = 60/150 + 55/150
        = 105/150
        = 7/10 .
      3. P(mango or vanila) = P(mango) + P(vanila)
        = 35/150 + 55/150
        = 90/150
        = 3/5 .
  2. Here is a case of NON-Exclusiveness. If some elements are share by event A and event B then
    P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
    where P(A ∩ B) is read as probability of event A AND event B. So P(A ∩ B) is there to avoid double-accounting for some elements both in A and B.
    1. Example 2: A bag contains 10 chips numbered from 1 to 10. A chip is selected from this bag at random.
      Let A be the event when the selected number is odd and B be the event when the selected number is a factor of 6.
      The probability of obtaining either an odd number or a factor of 6 can then be calculated as follow.
      P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
      Possibilities of A = { 1, 3, 5, 7, 9} and thus P(A)= 5/10. Possibilities of B = {1, 2, 3, 6} and P(B)=4/10. Obviously
      P(A ∪ B) ¹ P(A) + P(B) because 1 and 3 appear in both A and B. To avoid double accounting we need to deduct
      P(A ∩ B) =2/10 from our previous result. Thus,
      P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
      P(A ∪ B)= 5/10 + 4/10 -2/10
      P(A ∪ B)= 7/10.
      Of course, for small lists like above we could also count all possible outcomes. In this case { 1, 2, 3, 5, 6, 7, 9} seven out of ten possibilities.

    2. Example 3: If P(A)=0.7, P(B)=0.4 and P(A ∩ B)= 0.35. Find P(A ∪ B)
      Solution:
      P(A ∪ B)= P(A) + P(B) - P(A ∩ B)
      P(A ∪ B)= 0.7 + 0.4 - 0.35
      P(A ∪ B)= 0.75
      Questions on probability at this level can most often be solved using a diagram. An appropriate diagram here is the Venn Diagram. We represent the universe with a square. A group or set is represented by an enclosure that can takes any shape. Here we have an oval shape for A and a rectangle for B. Do not forget to label your sets as in here.
      i) Enter 0.35 into the intersection between A and B.
      ii) Calculate the remaining probability in B not in A. 0.4-0.35= 0.05.
      iii) The probability of A union B is just P(A)+P(B not in A)= 0.7 + 0.05 = 0.75.

      The Venn Diagram suggests that A and B are not exhaustive events because P( A∪ B ) ≠ 1.
      If events A, B and C are exhaustive events then the union of these events is the sample size and P(A∪ B ∪ C) = 1.
    3.  

    4. Example 4: In a class there are 20 students. 10 students play badminton and 15 students play soccer. 6 students play both badminton and soccer. One student did not play either badminton or soccer. A student is selected from this class. Find the probability that the student :
      1. plays either badminton or soccer
      2. plays badminton but not soccer
      3. plays soccer only
      Solutions: We enter the appropriate number of student for each region on the Venn Diagram. As usual, we start by entering the number that "carries the greatest number of information" into the intersection, that is 6 students play both badminton and soccer. "6" is entered into the intersection of these sets. Since there are only 10 students who play badminton then the rest of badminton set will receive "4." Similarly, the rest of soccer will receive "9." The one student who plays neither badminton nor soccer is represented by "1" outside these two sets. With these diagram, it is a matter of reading off the diagram to solve the above questions.
      1. P(badminton or soccer) = (4+6+9) / 20
          = 19/20
        Note that this is not simply P(badminton) + P(soccer) = 10/20 + 15/20 which is greater than 1!
        P(badminton or soccer)= P(badminton) + P(soccer) - P(badminton and soccer).
      2. P(badminton and not soccer) = 4/20 from the venn diagram.
      3. P(soccer only) = 9/20 from the diagram.

  3. We say that event A and event B are independent if
    P(A ∩ B)= P(A) x P(B).
    In IB, statistical independence carries the same meaning as independent as defined above.
    Note that A is independent to B does not suggest that P(A ∩ B) is zero.

  4. Independence and With Replacement.
    Example 5:
    A bag contains 5 yellow chips and 8 red chips. A chip is picked at random from the bag, the color is noted and the chip is replaced into the bag. Thus, there are 13 chips in the bag before every pick.
    1. What is the probability of obtaining 2 yellow chips on two picks?
    2. What is the probability of obtaining a yellow chip and a red chip?
    3. What is the probability of obtaining first a yellow chip and the a red chip?
    Solutions:
    Method 1:
    This is a case of replacement since every pick we have the same probability of obtaining say the yellow chip. When we have replacement, the probability of obtaining a yellow is independent of previous outcome. Thus, solving these question is a matter of applying
    P(A ∩ B)= P(A) x P(B).
    1. P(Y ∩ Y)= P(Y) x P(Y) = 5/13 x 5/13
        = 25/169
    2. Here we have two possibilities, we could have first yellow and then red or first red and then yellow.
      P(Y ∩ R) + P(R ∩ Y) = P(Y) x P(R) + P(R) x P(Y)
        = 2[ P(Y) x P(R)]
        = 2[5/13 x 8/13]
        = 80 / 169 .
    3. P(Y ∩ R) = P(Y) x P(R). In this case the order is important. Thus, the answer is 40/169.
    Method 2:
    You could also solve the above questions by drawing a similar tree diagram as shown here.
    1. The answer is obtained from the first branch on second pick. Answer is 5/13 x 5/13 = 25/169.
    2. Two possibilities so we need to add P(Y ∩ R) to P(R ∩ Y). Answer is 80/169.
    3. The answer is in the second branch on second pick. Answer is 5/13 x 8/13 = 40/169.
    Observe that since this is an experiment with replacement and the events are independent then P(Y) and P(R) on each pick are 5/13 and 8/13 respectively. The denominator for each probability is 13 because the chip is always replaced making the total number 13.

  5. Without Replacement.
    Example 6: Let us revisit our previous bag with 5 yellow chips and 8 red chips. This time, we will not replaced the selected chips.
    Thus, every pick will reduce the total number of chips in the bag by one.
    1. What is the probability of obtaining 2 yellow chips on two picks?
    2. What is the probability of obtaining a yellow chip and a red chip?
    3. What is the probability of obtaining first a yellow chip and the a red chip?
    Solutions:
    The best method do solve this question is to use a tree diagram. Note how the probability for each branch has been entered.
    1. P(Y ∩ Y)= P(Y) x P(Y)
        = 5/13 x 4/12
        = 20/156 = 10/78
      Note that the p(yellow on the second pick given the first pick was yellow) is 4/12 because a yellow is picked in the previous round and reduced the number of yellow chips from 5 to 4. At the same time, the total number of chips is reduced from 13 to 12 because the selected chip from the first pick was NOT replaced.
    2. Here there are two possibilities:
      P(Y ∩ R) + P(R ∩ Y)= P(Y) x P(R) + P(R) x P(Y)
        = [5/13 x 8/12 ] + [8/13 x 5/12]
        = 80/156 = 20/39.
      P(red on the second pick given that the first pick was yellow) = 8/12 because a yellow was picked in the first so number of red chips remained at 8. However, the total number is reduced by one because of no replacement. Similar argument applies to P(yellow on the second pick given the first pick was a red).
    3. P(Y ∩ R) = 5/13 x 8/12 = 10/39.

  6. Example 7: Grace and Junius will play a game. They will pick without replacement a marble at random from a bag. The bag contains 3 red marbles and 7 black marbles. The winner is the one who picked a red marble and the opponent picked a black marble. There is no winner if both picked the same colour.
    1. What is the probability that Grace will win if she picks first?
    2. What is the probability that Grace will win if she picks after Junius?
    3. What is the probability that there is no winner in this game?

    Solutions: The tree diagram for this question looks like this.
    1. If Grace is to win as the first to pick then the only outcome is RB.
      P(first red and second black) = 3/10 x 7/9 (second end branch) = 21/90.
    2. If Grace is to win as the second to pick then the only outcome is BR.
      P(first black and second red) = 7/10 x 3/9 (third end branch) = 21/90.
    3. No winner when the outcomes are RR or BB regardless who goes first.
      P(no winner) = P(RR) + P(BB)
        = [3/10 x 2/9] + [7/10 x 6/9]
        = 6/90 + 54/90
        = 60/90
        = 2/3.

Exercises