disjoint

Examples of disjoint in the following topics:

• Disjoint or mutually exclusive outcomes

  • (a) Are the outcomes none, small, and big disjoint?
  • The Addition Rule applies to both disjoint outcomes and disjoint events.
  • ( b) Are events B and D disjoint?
  • (c) Are events A and D disjoint?
  • This means they are disjoint outcomes.

• Probabilities when events are not disjoint

  • Events A and B are not disjoint { the cards J<>, Q<>, and K<>} fall into both categories <> so we cannot use the Addition Rule for disjoint events.
  • (a) If A and B are disjoint, describe why this implies P(A and B) = 0.
  • (b) Using part (a), verify that the General Addition Rule simplifies to the simpler Addition Rule for disjoint events if A and B are disjoint.10
  • 2.17: (a) If A and B are disjoint, A and B can never occur simultaneously.
  • (b) If A and B are disjoint, then the last term of Equation (2.16) is 0 (see part (a)) and we are left with the Addition Rule for disjoint events.

• Complement of an event

  • We use the Addition Rule for disjoint events to apply Property (ii):
  • 2.21: (a) The outcomes are disjoint and each has probability 1=6, so the total probability is 4/6 = 2/3. ( b) We can also see that P(D) = 1/6 + 1/6 = 1/3.
  • Since D and Dc are disjoint, P(D) + P(Dc) = 1.
  • (b) Noting that each outcome is disjoint, add the individual outcome probabilities to get P(Ac) = 2/3 and P(Bc) = 2/3.
  • (c) A and Ac are disjoint, and the same is true of B and Bc.

• The Addition Rule

  • Suppose $A$ and $B$ are disjoint, their intersection is empty.
  • These two events are disjoint, since there are no kings that are also queens.

• Probability distributions

  • A probability distribution is a table of all disjoint outcomes and their associated probabilities.

• Marginal and joint probabilities

  • Verify Table 2.14 represents a probability distribution: events are disjoint, all probabilities are non-negative, and the probabilities sum to 1.24.
  • 2.36: Each of the four outcome combination are disjoint, all probabilities are indeed non-negative, and the sum of the probabilities is 0.28 + 0.19 + 0.21 + 0.32 = 1.00.

• Fundamentals of Probability

  • If one event occurs in $30\%$ of the trials, a different event occurs in $20\%$ of the trials, and the two cannot occur together (if they are disjoint), then the probability that one or the other occurs is $30\% + 20\% = 50\%$.
  • Thus when $A$ and $B$ are disjoint, we have $P(A \cup B) = P(A)+P(B)$.
  • Then $A=\{HT,TH\}$ and $B=\{TT\}$ are disjoint.

• Unions and Intersections

  • If sets $A$ and $B$ are disjoint, however, the event $A \cap B$ has no outcomes in it, and is an empty set denoted as $\emptyset$, which has a probability of zero.
  • So, the above rule can be shortened for disjoint sets only:
  • This can even be extended to more sets if they are all disjoint:

• Defining probability exercises

  • (a) Are being Independent and being a swing voter disjoint, i.e. mutually exclusive?
  • (a) Are living below the poverty line and speaking a language other than English at home disjoint?
  • In parts (a) and (b), identify whether the events are disjoint, independent, or neither (events cannot be both disjoint and independent).
  • (d) Add up the corresponding disjoint sections in the Venn diagram: 0.24 + 0.11 + 0.12 = 0.47.
  • If graded on a curve, then neither independent nor disjoint (unless the instructor will only give one A, which is a situation we will ignore in parts (b) and (c).

• Probability solutions

  • (d) Add up the corresponding disjoint sections in the Venn diagram: 0.24 + 0.11 + 0.12 = 0.47.
  • If graded on a curve, then neither independent nor disjoint (unless the instructor will only give one A, which is a situation we will ignore in parts (b) and (c).
  • The other two orderings have the same probability, and these three possible orderings are disjoint events.