Free math worksheets

- Ask whether both events can happen on the same trial. - When combining exclusive events, think about whether any outcome would be counted twice. - Do not confuse exclusivity with independence.

1. The experiment records one outcome for a single card draw. 2. The descriptions “drawing a heart” and “drawing a club” cannot both hold for the same outcome. 3. Thus \(A\cap B=\varnothing\), so the events are mutually exclusive.

Yes. The events are mutually exclusive because \(A\cap B=\varnothing\).

- Use the addition rule for a union. - Mutual exclusivity makes the intersection probability \(0\). - Add the two given event probabilities after accounting for the overlap.

1. Mutually exclusive events have \(P(A\cap B)=0\). 2. Therefore \(P(A\cup B)=P(A)+P(B)=0.2+0.4=0.6\).

\(P(A\cup B)=0.6\).

- Identify which component of the ordered outcome each event constrains. - Different stages can carry different colors simultaneously. - Construct one shared ordered outcome.

1. The color conditions apply to different draw positions. 2. An outcome such as \((\text{red first}, \text{blue second})\) satisfies both. 3. Different labels are not mutually exclusive when they can occur at different stages.

The events are not mutually exclusive; “red first, blue second” lies in their intersection.

- Start with the possible values contained in the larger event. - Choose category boundaries so that every allowed value enters exactly one piece. - Check both overlap and coverage.

1. Use \(E_1=\{X=1\}\), \(E_2=\{X=2\}\), and \(E_3=\{X\ge3\}\). 2. No value of \(X\) can satisfy two of these descriptions, so the events are pairwise mutually exclusive. 3. Their union is \(\{X\ge1\}\).

One valid split is \(E_1=\{X=1\}\), \(E_2=\{X=2\}\), and \(E_3=\{X\ge3\}\). These events are pairwise mutually exclusive, and \(E_1\cup E_2\cup E_3=\{X\ge1\}\).

- Identify what one outcome permits. - The same labels can represent different event structures under different survey rules. - Look for whether a single respondent can satisfy both events.

1. In the exactly-one survey, a respondent cannot be in both categories, so the events are mutually exclusive. 2. In the select-all survey, one respondent may use both TV and online, so the events can overlap. 3. Mutual exclusivity depends on the outcome definition, not only the category names.

They are mutually exclusive in the exactly-one survey but not necessarily in the select-all survey.

- Check boundary values explicitly. - Mutually exclusive intervals cannot share an endpoint. - Exhaustive intervals must leave no gap.

1. Both events contain \(X=10\), so their intersection is not empty. 2. Replace one event with \(B=\{X>10\}\), or replace \(A\) with \(\{X<10\}\). 3. The revised pair has no overlap and still covers every real value.

No; they overlap at \(X=10\). One valid revision is \(A=\{X\le10\}\) and \(B=\{X>10\}\).

- One person occupies exactly one finishing position. - Use symmetry for each endpoint event. - Add the probabilities only after checking the intersection.

1. Maya cannot occupy both first and last place in one ordering, so \(A\cap B=\varnothing\). 2. By symmetry, \(P(A)=P(B)=\frac15\). 3. Therefore \(P(A\cup B)=\frac15+\frac15=\frac25\).

The events are mutually exclusive, and \(P(A\cup B)=\frac25\).

- Use possible score examples. - Events from different classification variables may overlap. - Do not transfer exclusivity from one partition to an unrelated event.

1. A home win can include a home goal. 2. A draw can include a positive tied score, so it can include a home goal. 3. An away win can also occur when the away team wins \(2\) to \(1\), so it can include a home goal. Thus \(G\) can overlap all three result events.

\(G\) can overlap home win, draw, and away win. Match-result categories are exclusive with one another, but not with the scoring event.

- Ask whether one category can be contained inside the other. - Describe an outcome that would satisfy both event descriptions. - Distinct names do not necessarily mean distinct, nonoverlapping sets.

1. Every wolf is a mammal, so \(A\subseteq B\). 2. Therefore \(A\cap B=A\), which is nonempty whenever wolves are possible. 3. Different verbal labels can describe nested events rather than disjoint categories.

The events are not mutually exclusive. Their intersection is \(A\cap B=A\), because every wolf is a mammal.

- Locate the values that satisfy both interval conditions. - For a uniform interval, compare lengths. - What intersection probability is required for mutual exclusivity?

1. Values satisfying both inequalities lie from \(4\) through \(6\), so \(A\cap B=[4,6]\). 2. The overlap has length \(2\) within a total interval of length \(10\). 3. Thus \(P(A\cap B)=\frac{2}{10}=0.20\), which is positive, so the events are not mutually exclusive.

a) \(A\cap B=[4,6]\) b) \(P(A\cap B)=0.20\) c) The events are not mutually exclusive.

- Check each of the three pairwise intersections separately. - A transitive relation would force the first and third events to have the relation. - One shared outcome is enough to refute mutual exclusivity.

1. \(A\cap B=\varnothing\), so \(A\) and \(B\) are mutually exclusive. 2. \(B\cap C=\varnothing\), so \(B\) and \(C\) are mutually exclusive. 3. \(A\cap C=\{2\}\ne\varnothing\), so \(A\) and \(C\) are not mutually exclusive. 4. Therefore, mutual exclusivity does not satisfy the transitive property.

\(A\cap B=\varnothing\) and \(B\cap C=\varnothing\), but \(A\cap C=\{2\}\). Thus mutual exclusivity is not transitive.

- Expand the phrase “within \(20^\circ\) of \(0^\circ\)” into intervals. - Remember that \(0^\circ\) and \(360^\circ\) represent the same direction. - Compare the second interval with event \(B\).

1. The interval \(B=[350^\circ,355^\circ]\) lies inside the wrap-around part \([340^\circ,360^\circ)\) of \(A\). 2. Therefore \(A\cap B=B\), which is nonempty. 3. On a circle, angles near \(360^\circ\) are also near \(0^\circ\); a linear reading that ignores wrap-around would miss this overlap.

No. \(A\cap B=[350^\circ,355^\circ]\). Circular wrap-around makes angles near \(360^\circ\) part of the neighborhood of \(0^\circ\).

- Translate minimum and maximum conditions into ordered outcomes. - Both conditions can determine the two values simultaneously. - Include both possible orders.

1. Both conditions hold when the two results are \(2\) and \(5\) in either order. 2. \(A\cap B=\{(2, 5),(5, 2)\}\). 3. Therefore the events are not mutually exclusive.

No. Their intersection is \(\{(2, 5),(5, 2)\}\).

- Assign at least one distinct outcome to every nonempty event. - Use the fact that disjoint events cannot share an outcome. - Test the family of singleton events.

1. Each nonempty event in a pairwise disjoint family must contain at least one elementary outcome. 2. No elementary outcome can belong to two events in the family. 3. With only \(7\) outcomes, there can be at most \(7\) nonempty pairwise disjoint events. 4. The seven singleton events attain the bound.

The maximum is \(7\), attained by the seven singleton events.

- Construct one pair with no overlap but an incomplete union. - Construct another pair whose union is full but whose intersection is nonempty. - A fair die offers a small sample space for checking both properties.

1. On a die, “roll \(1\)” and “roll \(2\)” are mutually exclusive but do not cover all outcomes. 2. “Roll even” and “roll at most \(5\)” are exhaustive because outcomes \(1\) through \(5\) satisfy the second event and outcome \(6\) satisfies the first. They overlap at \(2\) and \(4\). 3. Thus the two properties are logically distinct.

Example: \(\{1\}\) and \(\{2\}\) are mutually exclusive but not exhaustive. “Roll even” and “roll at most \(5\)” are exhaustive but not mutually exclusive.

- Translate “both” into an intersection before negating it. - De Morgan’s law changes a complemented intersection into a union of complements. - Compare the resulting event with the original intersection.

1. “Both sensors exceed” is \(A\cap B\). 2. “At least one does not exceed” is its complement: \((A\cap B)^c\). 3. De Morgan’s law gives \((A\cap B)^c=A^c\cup B^c\). 4. An event and its complement are mutually exclusive and exhaustive, so \(A^c\cup B^c\) is mutually exclusive with \(A\cap B\).

The event is \(A^c\cup B^c=(A\cap B)^c\). It is mutually exclusive with \(A\cap B\).

- Where can event \(B\) occur if no outcome may belong to both events? - Compare \(B\) with the complement of \(A\). - Think about what must happen for the available probability outside \(A\) to be used completely.

1. Since \(A\cap B=\varnothing\), event \(B\) must lie within \(A^c\). 2. \(P(A^c)=1-0.62=0.38\), so \(P(B)\le0.38\). 3. Equality occurs when \(B=A^c\), so the two events are also exhaustive.

a) The greatest possible value is \(0.38\). b) It is attained when \(B=A^c\), making \(A\) and \(B\) exhaustive as well as mutually exclusive.

- A subset disjoint from \(B\) cannot contain any element of \(B\). - Start with all elements of \(A\) and remove exactly those that cause overlap. - To prove maximality, examine every element of \(A\) left outside the proposed set.

1. Any element of \(C\) must lie in \(A\) but not in \(B\), so \(C\subseteq A\setminus B\). 2. \(A\setminus B=\{1,2,3\}\), and this set has empty intersection with \(B\). 3. Taking all elements of \(A\setminus B\) gives the largest possible disjoint subset. Adding either \(4\) or \(5\), the remaining elements of \(A\), would create an intersection with \(B\).

The largest subset is \(C=A\setminus B=\{1,2,3\}\). It is maximal because every element of \(A\) not in \(C\) belongs to \(B\).

- Describe what remains after excluding each of two partition categories. - Use both disjointness and exhaustiveness of the original partition. - Consider where a third category lies relative to the two complements.

1. Because \(A,B,C\) are disjoint and exhaustive, outcomes outside both \(A\) and \(B\) are exactly the outcomes in \(C\). 2. Thus \(A^c\cap B^c=C\). 3. Since \(P(C)>0\), the complements overlap and are not mutually exclusive. 4. In any partition with at least three nonempty cells, the complements of two cells share every other cell.

a) \(A^c\cap B^c=C\) b) No, because their intersection has positive probability. c) The complements of two partition cells overlap on all remaining nonempty cells.

- A union probability cannot exceed \(1\). - Use the general addition rule rather than the disjoint version. - Maximize the union to minimize the overlap.

1. If the events were mutually exclusive, their union would have probability \(0.72+0.46=1.18\), which is impossible. 2. The addition rule gives \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\). 3. Since \(P(A\cup B)\le1\), the intersection is at least \(1.18-1=0.18\).

They cannot be mutually exclusive. The smallest possible intersection probability is \(0.18\).

- Exhaustive and mutually exclusive events assign each outcome to exactly one side. - Count binary assignments of the elementary outcomes. - Remove assignments that leave one labeled event empty.

1. If \(A\) and \(B\) are disjoint and exhaustive, every outcome must be assigned to exactly one of the two labeled events. 2. There are \(2^5=32\) assignments to A or B. 3. Exclude the assignment sending every outcome to A and the assignment sending every outcome to B, because both events must be nonempty. 4. The number is \(32-2=30\).

\(30\) ordered pairs.

- Check every pair, not only the intersection of all three. - Pairwise mutual exclusivity requires three separate empty intersections. - A triple intersection can be empty even when pairs overlap.

1. \(A\cap B=\varnothing\). 2. \(A\cap C=\{1\}\) and \(B\cap C=\{3\}\). 3. An empty triple intersection does not imply pairwise mutual exclusivity; two pairs overlap.

No. Only \(A\) and \(B\) are mutually exclusive; \(C\) overlaps both.

- Separate evidence from a finite sample from logical impossibility. - Ask whether the category definitions prohibit overlap. - Zero frequency and structural zero are not automatically the same.

1. A sample zero means no overlap appeared in the observed data. 2. Mutual exclusivity is a property of the outcome definitions in the population. 3. The overlap is structurally impossible only if licensing rules make the combination impossible; the count alone does not prove that.

No. An observed zero may be a sample zero. Mutual exclusivity requires the combination to be impossible by definition or rule.

- Add the lower bounds for all disjoint events. - Compare the resulting sum with the maximum possible probability. - Check that an example reaches the integer bound.

1. Pairwise mutually exclusive probabilities add, so \(0.22k\le1\). 2. This gives \(k\le4.545\ldots\), hence \(k\le4\). 3. Four events are possible, for example with probabilities \(0.22,0.22,0.22,\) and \(0.34\). 4. Their probabilities sum to \(1\), so the maximum is \(4\).

The greatest possible value is \(k=4\). One attainable set of probabilities is \(0.22,0.22,0.22,0.34\).

- Count the possible yes/no patterns for three attributes. - Represent each pattern with an intersection of events or complements. - Check that every response has exactly one pattern.

1. Each attribute can be present or absent, giving \(2^3=8\) truth-value patterns. 2. The categories are \(A\cap B\cap C\), \(A\cap B\cap C^c\), \(A\cap B^c\cap C\), \(A\cap B^c\cap C^c\), \(A^c\cap B\cap C\), \(A^c\cap B\cap C^c\), \(A^c\cap B^c\cap C\), and \(A^c\cap B^c\cap C^c\). 3. Each response pattern belongs to exactly one of these categories, so they are pairwise disjoint and exhaustive. 4. No coarser standard scheme preserves all eight distinct yes/no patterns.

The \(8\) categories are \(A\cap B\cap C\), \(A\cap B\cap C^c\), \(A\cap B^c\cap C\), \(A\cap B^c\cap C^c\), \(A^c\cap B\cap C\), \(A^c\cap B\cap C^c\), \(A^c\cap B^c\cap C\), and \(A^c\cap B^c\cap C^c\).

- Write “exactly one” using set differences. - Ask what the set differences become when there is no overlap. - Once the event is identified, use the consequence of disjointness for its probability.

1. In general, \(A\triangle B=(A\setminus B)\cup(B\setminus A)\). 2. If \(A\cap B=\varnothing\), then \(A\setminus B=A\) and \(B\setminus A=B\). 3. Therefore \(A\triangle B=A\cup B\). 4. The disjoint-union probability is \(0.22+0.31=0.53\).

a) \(A\triangle B=A\cup B\) b) \(P(A\triangle B)=0.53\)

- One piece contains the outcomes of \(A\) that are in \(B\); the other contains those outside \(B\). - Use \(B\cap B^c=\varnothing\) and \(B\cup B^c=S\). - Apply additivity only after proving the pieces are disjoint.

1. \(C\cap D=A\cap B\cap A\cap B^c=A\cap(B\cap B^c)=\varnothing\), so \(C\) and \(D\) are mutually exclusive. 2. \(C\cup D=(A\cap B)\cup(A\cap B^c)=A\cap(B\cup B^c)=A\). 3. By additivity for disjoint events, \(P(A)=P(A\cap B)+P(A\cap B^c)\).

a) \(C\cap D=\varnothing\). b) \(C\cup D=A\). c) \(P(A)=P(A\cap B)+P(A\cap B^c)\).

- Use the addition rule to find the union. - The complement of the union is the “neither” region. - Check all four regions and their total.

1. The addition rule gives \(P(A\cup B)=0.60+0.40-0.15=0.85\). 2. Therefore the probability of neither event is \(1-0.85=0.15\). 3. One valid four-region model assigns \(0.15\) to both, \(0.45\) to A only, \(0.25\) to B only, and \(0.15\) to neither. 4. Marginal probabilities summing to \(1\) do not force the overlap and neither regions to be zero.

A valid model is: both \(0.15\), A only \(0.45\), B only \(0.25\), neither \(0.15\). The events are neither mutually exclusive nor exhaustive.

- Use the fact that both new events stay inside the original event being split. - Check pairwise overlap separately from total coverage. - Replace the union of the two new pieces with the original event.

1. Because \(B_1,B_2\subseteq B\), each is disjoint from \(A\) and \(C\), which were disjoint from \(B\). 2. \(B_1\) and \(B_2\) are disjoint by assumption. 3. The union is \(A\cup B_1\cup B_2\cup C=A\cup B\cup C=S\). 4. The four events are pairwise mutually exclusive and exhaustive, so they form a partition.

The four events form a partition: they are pairwise mutually exclusive, and \(A\cup B_1\cup B_2\cup C=A\cup B\cup C=S\).

- Translate mutual exclusivity into a statement about individual outcomes. - What does membership in a complement mean? - Prove each implication separately rather than assuming they are reversible.

1. If \(A\cap B=\varnothing\), then no outcome in \(A\) belongs to \(B\); therefore every outcome in \(A\) belongs to \(B^c\), so \(A\subseteq B^c\). 2. If \(A\subseteq B^c\), an outcome in \(A\) cannot also be in \(B\); therefore \(A\cap B=\varnothing\). 3. The two statements are equivalent.

\(A\cap B=\varnothing\) if and only if \(A\subseteq B^c\). The first condition says no outcome lies in both events, which is exactly the requirement that every outcome of \(A\) lie outside \(B\).

- Use the logical implication built into the primary-rail category. - Events from primary and secondary travel descriptions need not form one partition. - Test each pair separately.

1. If the primary route is rail, the commuter uses a train, so \(R\cap T^c=\varnothing\). 2. A highway-primary commuter may use a train for part of the trip under the stated definition. 3. Therefore highway and \(T\) are not necessarily mutually exclusive.

\(R\) and \(T^c\) are mutually exclusive. Highway and \(T\) may overlap.

- Check what was removed from each later-defined set. - Test each pairwise intersection. - Track where an outcome belonging to several original events is assigned.

1. \(D_2\) contains no outcomes from \(A=D_1\), so \(D_1\cap D_2=\varnothing\). 2. \(D_3\) removes every outcome in \(A\cup B\), so it overlaps neither \(D_1\) nor \(D_2\). 3. Every outcome in \(A\cup B\cup C\) is assigned to the first event in the order \(A,B,C\) that contains it. 4. Thus the three disjoint pieces have the same union as the original events and can be counted without overlap corrections.

\(D_1,D_2,D_3\) are pairwise disjoint and their union is \(A\cup B\cup C\). They partition the union into nonoverlapping pieces, so their counts or probabilities can be added directly.

- Use contradiction: suppose the intersection contains one elementary outcome. - Compare the probability of a set with the probability of a positive-probability subset it contains. - Distinguish an empty event from a nonempty null event.

1. If \(A\cap B\) contained an elementary outcome \(s_i\), then \(P(A\cap B)\ge P(\{s_i\})>0\), contradicting \(P(A\cap B)=0\). 2. Therefore \(A\cap B=\varnothing\), so the events are mutually exclusive. 3. Without the positive-atom assumption, a nonempty event may have probability \(0\), as can occur for single points in a continuous model. Then zero intersection probability would not force an empty intersection.

Under the stated finite positive-atom condition, \(A\cap B=\varnothing\), so \(A\) and \(B\) are mutually exclusive. The conclusion can fail when nonempty events of probability \(0\) are allowed.

- Expand the intersection of the two unions using distributivity. - Use mutual exclusivity to eliminate one term. - Any intersection with \(C\) is already contained in \(C\).

1. Distribute intersection over union: \((A\cup C)\cap(B\cup C)=(A\cap B)\cup(A\cap C)\cup(C\cap B)\cup C\). 2. Since \(A\cap B=\varnothing\), the first term disappears. 3. Both \(A\cap C\) and \(B\cap C\) are subsets of \(C\), so their union with \(C\) is simply \(C\).

\((A\cup C)\cap(B\cup C)=C\).