An archive the questions from Mark's Fall 2018 Stat 225.

Probabilty of sets

Mark

Suppose that A, B, and C are events with

  • P(A) = 0.22,
  • P(B) = 0.25,
  • P(C) = 0.28,
  • P(A\cap B) = 0.11,
  • P(A\cap C) = 0.05,
  • P(B\cap C) = 0.07, and
  • P(A\cap B \cap C) = 0.01.

Compute the probabilities of the following events and interpret them as statements:

  1. A \cup B
  2. A \cup B \cup C
  3. A^c \cap B^c
  4. A^c \cap B^c \cap C^c
megan

  1. The union of A and B is .36
    To find the union of A and B you must add the probabilities of A and B and then subtract the probability of the intersection of A and B. You must subtract the intersection so that it isn’t counted twice. In other words,

    .25 + .22 - .11 = .36

  2. The union of A, B and C is .53
    To find the union of A, B, and C you must add the probabilities of A, B, and C and then subtract the probability of the intersections of A and B, A and C, and B and C. Then you must add back the probability of the intersection of A, B and C. You must subtract the intersections of A and B, A and C, and B and C so that they aren’t counted twice and you must add back the intersection of all three because you have now taken it away. In other words,

    .22 + .25 + .28 = .75
    .75 - .11 - .05 - .07 = .52
    .52 + .01 = .53

  3. The intersection of A compliment and B compliment is .64
    The intersection of A compliment and B compliment is equivalent to the compliment of the union of A and B. Therefore, I just took my answer from part one, .36., and subtracted that from one. The compliment of the union of A and B just means everything except the union of A and B. In other words,

    1 - .36 = .64

  4. The intersection of the compliments of A, B, and C is .47
    Similar to how I did part 3, for this one I just took my answer from part two, .35, and subtracted that from 1. The intersection of the compliments of A, B, and C is equivalent to the compliment of the union of A, B, and C. The compliment of the union of A, B, and C just means everything except the union of A, B, and C. In other words.

    1 - .53 = .47

joshua
  1. P( A∪B) = P(A)+P(B)-P(A∩B)
    P( A∪B) = 0.22+0.25-0.11
    P( A∪B) = 0.36
  2. P( A∪B∪C) = P(A)+P(B)+P©-P(A∩B)-P(A∩C)-P(B∩C)+P(A∩B∩C)
    P( A∪B∪C) = 0.22+0.25+0.28-0.11-0.05-0.07+0.01
    P( A∪B∪C) = 0.53
  3. P( A^c∩B^c) = 1-P( A∪B)
    P( A^c∩B^c) = 1-(P(A)+P(B)-P(A∩B))
    P( A^c∩B^c) = 1-(0.36) Taken from Question 1
    P( A^c∩B^c) = 0.64
  4. P( A^c∩B^c∩C^c) = 1-P( A∪B∪C)
    P( A^c∩B^c∩C^c) = 1-(P(A)+P(B)+P©-P(A∩B)-P(A∩C)-P(B∩C)+P(A∩B∩C))
    P( A^c∩B^c∩C^c) = 1-(0.53) Taken from Question 2
    P( A^c∩B^c∩C^c) = 0.47
Mark

@joshua Have a look at this post for pointers on how to include really nicely typeset math in this forum!

Mark

I think we’ve got a couple of pretty good answers above but not quite full credit. I think that @megan’s answer is a bit too wordy and should be written symbolically. @joshua’s answer is better but could be written more as mathematical sentences. For example, I’d write the solution to number 2 as:

\begin{align} P(A \cup B \cup C) &= P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \\ &= 0.22 + 0.25 + 0.28 - 0.11 - 0.05 - 0.07 + 0.01 = 0.53. \end{align}
dennis

  1. P(AuuB) = 0.22 + 0.25 -.11 = 0.36
    The probability of A or B happening, (must subtract overlap).

  2. P(AuuBuuC) = 0.22 + .025 + 0.28 - .11 - .05 -.07 + .01 + = 0.53
    The probability of A or B or C happening (minus overlap).

  3. A^c is the compliment of A (had to google that…)
    1-A :.
    P(A^c) = 1- .22 =.78
    P(B^c) = 1- .25 =.75
    P(C^c) = 1- .28 =.72

P(A^c nn B^c) = 1-P(A uu B) = 1- .36 =0.64

P(A^c nn B^c nn C^c) = 1-P(A uu B uu C) = 1- 0.53 = .47