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Independent events and unions

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54725412
A fair die is rolled many times. A student says, “Because the last \(3\) rolls were all even, an odd result is now more likely.” Evaluate this statement using the probability model for repeated independent rolls.

Hints

- Focus on the probability model for the next roll. - Decide whether previous rolls change the six possible outcomes or their probabilities. - Count the odd faces on one fair die.

Solution

1. Each new fair-die roll has the same six equally likely outcomes regardless of previous rolls. 2. The probability of an odd result on the next roll is \(\frac{3}{6}=\frac12\). 3. A short run of previous results does not change the model for the next independent roll.

Answer

The statement is false. The next roll is still odd with probability \(\frac{1}{2}\).
54735612
Events \(A\) and \(B\) are mutually exclusive, with \(P(A)=0.25\) and \(P(B)=0.40\). Can they also be independent? Explain numerically.

Hints

- Compare the intersection required by each property. - Nonzero event probabilities make their product positive. - Two definitions can impose incompatible conditions.

Solution

1. Mutual exclusivity gives \(P(A\cap B)=0\). 2. Independence would require \(P(A\cap B)=0.25\cdot0.40=0.10\). 3. The requirements conflict, so the events cannot be independent.

Answer

No. Disjointness requires intersection \(0\), while independence would require \(0.10\).
54735812
Events \(A\) and \(B\) are independent, and \(P(A)=0.30\). Find the odds in favor of \(A\) among outcomes in \(B\), assuming \(P(B)>0\). Explain why the answer does not require the value of \(P(B)\).

Hints

- Independence leaves one event’s probability unchanged after conditioning on the other. - Convert a probability and its complement into odds. - Check whether the size of the conditioning event appears in the final ratio.

Solution

1. Independence gives \(P(A\mid B)=P(A)=0.30\). 2. Within \(B\), the complementary probability is \(P(A^c\mid B)=0.70\). 3. The conditional odds in favor of \(A\) are \(0.30\) to \(0.70\), or \(3\) to \(7\). 4. The value of \(P(B)\) cancels because conditioning on \(B\) does not change the probability of \(A\).

Answer

The odds in favor of \(A\) within \(B\) are \(3\) to \(7\).
54736412
Two tokens are drawn from the urn shown without replacement. Are the events “first token is red” and “second token is red” independent? Show using conditional probability.
Figure for problem 547364

Hints

- Compare a marginal probability with the corresponding conditional probability. - Update the bag after the first draw. - Without replacement changes the composition.

Solution

1. \(P(\text{second red})=\frac{4}{10}=0.40\) by symmetry. 2. \(P(\text{second red}\mid\text{first red})=\frac39=\frac13\). 3. Since \(\frac13\ne0.40\), the events are not independent.

Answer

No. The first red draw lowers the second-red probability from \(0.40\) to \(\frac13\).
54736512
A token is drawn from the urn shown, replaced, and the urn is remixed before a second draw. Are the events “first token is red” and “second token is red” independent?
Figure for problem 547365

Hints

- Track whether the first draw changes the second draw’s sample space. - Replacement restores the original probabilities. - Use the unchanged-conditional criterion.

Solution

1. Each draw has red probability \(\frac{4}{10}=0.40\). 2. Replacement restores the original composition, so \(P(\text{second red}\mid\text{first red})=0.40\). 3. The conditional equals the marginal, so the events are independent.

Answer

Yes. Replacement keeps the second-draw red probability at \(0.40\).
54736912
A student says two events are independent because one concerns favorite music and the other concerns shoe size. Explain why subject-matter unrelatedness is not enough to establish probabilistic independence. State what data or model relationship must be checked.

Hints

- Distinguish everyday meaning from the mathematical definition. - Look for a condition involving joint and marginal probabilities. - Possible confounding groups can create associations.

Solution

1. Independence is a numerical property of a joint distribution, not a judgment about topics. 2. One must verify \(P(A\cap B)=P(A)P(B)\), or an equivalent unchanged-conditional relationship. 3. Hidden population structure can associate variables that appear conceptually unrelated.

Answer

Conceptual unrelatedness does not prove independence. Check the joint-product or unchanged-conditional criterion in the model or data.
54737312
A point \((X,Y)\) is chosen uniformly from the displayed unit square. Let \(A=\{X<0.30\}\) and \(B=\{Y>0.60\}\). Use areas in the display to show that \(A\) and \(B\) are independent.
Figure for problem 547373

Hints

- Use strip widths as marginal probabilities. - The overlap is a rectangle. - Compare its area with the product of the strip areas.

Solution

1. \(P(A)=0.30\) and \(P(B)=0.40\). 2. Their intersection is a rectangle of area \(0.30\cdot0.40=0.12\). 3. Since \(P(A\cap B)=P(A)P(B)\), the events are independent.

Answer

The intersection area is \(0.12=0.30\cdot0.40\), so the events are independent.
54738412
A study reports \(P(A)=0.36\), \(P(B)=0.50\), and \(P(A\mid B)=0.44\). Are the events independent? Quantify the change caused by conditioning.

Hints

- Compare the conditional and marginal probabilities for the same event. - Equality is required for independence. - State both direction and size of the difference.

Solution

1. Independence would require \(P(A\mid B)=P(A)=0.36\). 2. The observed conditional probability is \(0.44\). 3. Conditioning increases the probability by \(0.08\), so the events are not independent.

Answer

No. Conditioning raises the probability from \(0.36\) to \(0.44\), a change of \(0.08\).
54734212
Six distinct books are arranged uniformly on a shelf. Given that book \(A\) appears before book \(B\), find the probability that book \(C\) appears before book \(D\).

Hints

- Conditioning fixes one pair’s relative order. - Use symmetry for the other pair. - Pairwise relative order can be counted without listing all \(6!\) arrangements.

Solution

1. The relative orders of the disjoint pairs \((A, B)\) and \((C, D)\) are symmetric. 2. Among arrangements with \(A\) before \(B\), exactly half have \(C\) before \(D\). 3. The conditional probability is \(\frac12\).

Answer

The probability is \(\frac12\).
54734512
Two fair coins are flipped independently. Let \(H_1\) and \(H_2\) be the events that the first and second coins show heads, and let \(C=H_1\cup H_2\) be the event “at least one head.” Show that \(H_1\) and \(H_2\) are not conditionally independent given \(C\), even though they are independent before conditioning.

Hints

- Restrict the ordered coin sample space to outcomes satisfying the condition. - Compute both conditional marginals and the conditional intersection. - Compare the intersection with the product criterion inside the restricted space.

Solution

1. Given \(C\), the equally likely outcomes are \(HH, HT, TH\). 2. \(P(H_1\mid C)=\frac23\) and \(P(H_2\mid C)=\frac23\). 3. \(P(H_1\cap H_2\mid C)=\frac13\). 4. Since \(\frac13\ne\frac23\cdot\frac23=\frac49\), conditional independence fails. 5. Knowing one coin is tails under the “at least one head” condition forces the other coin to be heads, creating dependence.

Answer

The events are not conditionally independent given \(C\): \(P(H_1\cap H_2\mid C)=\frac13\), while \(P(H_1\mid C)P(H_2\mid C)=\frac49\).
54735312
Two probability models use binary variables \(X,Y\in\{0,1\}\). In both models, each variable is fair: \(P(X=1)=P(Y=1)=\frac12\). Model I assigns probability \(\frac14\) to each ordered pair. Model II assigns probability \(\frac12\) to \((0, 0)\), probability \(\frac12\) to \((1, 1)\), and probability \(0\) to the other pairs. Compare independence in the two models and explain why identical marginals do not determine the joint relationship.

Hints

- Check a joint cell against the product of its marginal probabilities. - Compare where the probability mass is placed in the two tables. - Separate information about each variable alone from information about their pairing.

Solution

1. In Model I, \(P(X=1,Y=1)=\frac14=\frac12\cdot\frac12\), and the full joint table factors, so \(X\) and \(Y\) are independent. 2. In Model II, \(P(X=1,Y=1)=\frac12\ne\frac14\), so the variables are dependent; in fact \(Y=X\) always. 3. Both models have the same marginal distributions, but their probability mass is arranged differently across joint outcomes.

Answer

Model I is independent. Model II is dependent even though the marginals are the same. Marginals alone do not determine a joint distribution.
54736712
Independent events \(A\) and \(B\) have the same probability \(p\). The probability that exactly one occurs is \(0.48\). Find all possible values of \(p\).

Hints

- Write the two symmetric one-event cases. - Combine them into one expression in \(p\). - Both roots can be valid probabilities.

Solution

1. Exactly one has probability \(2p(1-p)\). 2. Solve \(2p(1-p)=0.48\), giving \(p^2-p+0.24=0\). 3. The roots are \(p=0.40\) and \(p=0.60\).

Answer

\(p=0.40\) or \(p=0.60\).
54736812
A probability space has four elementary outcomes with probabilities \(P(\omega_1)=0.10\), \(P(\omega_2)=0.20\), \(P(\omega_3)=0.30\), and \(P(\omega_4)=0.40\). Let \(A=\{\omega_1,\omega_2\}\). Can there be an event \(B\) with \(P(B)=0.50\) that is independent of \(A\)? Prove your conclusion by considering every possible value of \(P(A\cap B)\).

Hints

- First determine the probability independence would require for the intersection. - The intersection can include only elementary outcomes already inside \(A\). - List the subset probabilities available inside \(A\) and compare them with the required value.

Solution

1. \(P(A)=0.10+0.20=0.30\). 2. If \(A\) and \(B\) were independent with \(P(B)=0.50\), then \(P(A\cap B)=0.30\cdot0.50=0.15\). 3. The intersection \(A\cap B\) can contain neither outcome of \(A\), only \(\omega_1\), only \(\omega_2\), or both. Its possible probabilities are therefore \(0\), \(0.10\), \(0.20\), and \(0.30\). 4. Since \(0.15\) is not possible, no such event \(B\) exists.

Answer

No. Independence would require \(P(A\cap B)=0.15\), but an intersection with \(A\) can have probability only \(0\), \(0.10\), \(0.20\), or \(0.30\).
54737012
Five fair coin flips are independent. Let \(A\) be “the first two flips match” and \(B\) be “the last three flips contain exactly one head.” Prove that \(A\) and \(B\) are independent without listing all \(32\) outcomes.

Hints

- Identify which random trials each event uses. - Compute each block event within its own smaller sample space. - Combine independent blocks rather than enumerating the full experiment.

Solution

1. Event \(A\) depends only on flips \(1\) and \(2\), while \(B\) depends only on flips \(3,4,5\). 2. The two blocks of flips are independent. 3. \(P(A)=\frac24=\frac12\) and \(P(B)=\frac{\binom31}{2^3}=\frac38\). 4. The joint event combines any successful first block with any successful second block, so \(P(A\cap B)=\frac12\cdot\frac38=\frac{3}{16}\). 5. Therefore \(A\) and \(B\) are independent.

Answer

\(P(A)=\frac12\), \(P(B)=\frac38\), and \(P(A\cap B)=\frac{3}{16}\). The events are independent because they depend on disjoint blocks of independent flips.
54737112
Events \(A,B,C\) are mutually independent, with \(P(A)=0.40\) and \(P(B\cap C)>0\). Find \(P(A\mid B\cap C)\). Show the calculation from the definition rather than citing the result only.

Hints

- Write the conditional probability as a ratio involving a triple intersection. - Use mutual independence in both the numerator and denominator. - Identify the factors that cancel.

Solution

1. Mutual independence gives \(P(A\cap B\cap C)=P(A)P(B)P(C)\). 2. It also gives \(P(B\cap C)=P(B)P(C)\). 3. Therefore \(P(A\mid B\cap C)=\frac{P(A)P(B)P(C)}{P(B)P(C)}=P(A)=0.40\).

Answer

\(P(A\mid B\cap C)=0.40\). Conditioning on the joint occurrence of \(B\) and \(C\) does not change the probability of \(A\) under mutual independence.
54737212
Each component in the two displayed systems works independently with probability \(0.90\). a) Find the reliability of system a). b) Find the reliability of system b). c) Compare the reliabilities.
Figure for problem 547372

Hints

- Translate each system rule into an event. - Series uses simultaneous success; parallel is easier through simultaneous failure. - Compare the final probabilities on the same scale.

Solution

1. Series reliability is \(0.90^2=0.81\). 2. Parallel failure requires both components to fail: \(0.10^2=0.01\). 3. Parallel reliability is \(1-0.01=0.99\), which is \(0.18\) higher.

Answer

a) \(0.81\) b) \(0.99\) The parallel system is \(0.18\) more reliable.
54737412
Independent events \(A\) and \(B\) have probabilities \(0.40\) and \(0.25\). Given that exactly one of the two events occurred, find the probability that it was \(A\).

Hints

- Split “exactly one” into its two disjoint possibilities. - Identify which of those possibilities is favorable. - Normalize by the total probability of the restricted event.

Solution

1. \(P(A\cap B^c)=0.40\cdot0.75=0.30\). 2. \(P(A^c\cap B)=0.60\cdot0.25=0.15\). 3. The conditioning event “exactly one” has probability \(0.30+0.15=0.45\). 4. The conditional probability is \(\frac{0.30}{0.45}=\frac23\).

Answer

The probability is \(\frac23\).
54737512
Three independent events have probabilities \(0.20\), \(0.50\), and \(0.70\). Find the probabilities of none, exactly one, exactly two, and all three occurring.

Hints

- Partition outcomes by the number of occurring events. - For each pattern, multiply event or complement probabilities. - Add disjoint patterns within each count category and verify a total of \(1\).

Solution

1. None: \(0.80\cdot0.50\cdot0.30=0.12\). 2. Exactly one: \(0.20\cdot0.50\cdot0.30+0.80\cdot0.50\cdot0.30+0.80\cdot0.50\cdot0.70=0.03+0.12+0.28=0.43\). 3. Exactly two: \(0.20\cdot0.50\cdot0.30+0.20\cdot0.50\cdot0.70+0.80\cdot0.50\cdot0.70=0.03+0.07+0.28=0.38\). 4. All three: \(0.20\cdot0.50\cdot0.70=0.07\). 5. The four probabilities sum to \(1\).

Answer

None: \(0.12\) Exactly one: \(0.43\) Exactly two: \(0.38\) All three: \(0.07\)
54737912
Events \(A\) and \(B\) are independent and satisfy \(P(A\cup B)=1\). Prove that at least one of the events has probability \(1\).

Hints

- Substitute the independence product into the union formula. - Move all terms to expose a factorization. - A product of two nonnegative factors is zero only if one factor is zero.

Solution

1. Independence gives \(P(A\cap B)=P(A)P(B)\). 2. The union formula becomes \(1=P(A)+P(B)-P(A)P(B)\). 3. Rearranging gives \((1-P(A))(1-P(B))=0\). 4. Therefore \(P(A)=1\) or \(P(B)=1\).

Answer

At least one of \(P(A)\) or \(P(B)\) equals \(1\).
54738212
Events \(A\) and \(B\) have \(P(A)=0.50\), \(P(B)=0.40\), and \(P(A\cup B)=0.70\). Determine whether they are independent.

Hints

- Recover the intersection from the union. - Compute the product of the marginals independently. - Compare the two values.

Solution

1. The addition rule gives \(P(A\cap B)=0.50+0.40-0.70=0.20\). 2. The product \(P(A)P(B)=0.50\cdot0.40=0.20\). 3. The values match, so the events are independent.

Answer

Yes. The inferred intersection \(0.20\) equals \(P(A)P(B)\).
54738312
Two fair six-sided dice are rolled. Let \(A\) be “the first die is even” and \(B\) be “the sum is even.” Determine whether \(A\) and \(B\) are independent, even though event \(B\) uses the first die in its definition.

Hints

- Classify die results by parity rather than listing all \(36\) outcomes. - What parity must the second die have when both event conditions hold? - Compare the joint probability with the product of the marginals.

Solution

1. \(P(A)=\frac{18}{36}=\frac12\). 2. A sum is even when both dice have the same parity, so \(P(B)=\frac{18}{36}=\frac12\). 3. For \(A\cap B\), the first die is even and the sum is even, so the second die must also be even. There are \(3\cdot3=9\) outcomes, giving \(P(A\cap B)=\frac{9}{36}=\frac14\). 4. Since \(\frac14=\frac12\cdot\frac12\), the events are independent.

Answer

The events are independent because \(P(A\cap B)=\frac14=P(A)P(B)\).
54738512
A uniform sample space has \(12\) outcomes. Event \(A\) contains \(6\) outcomes. Construct an event \(B\) containing \(4\) outcomes that is independent of \(A\). State how many outcomes \(A\cap B\) must contain.

Hints

- Convert event sizes to probabilities. - Use the product condition to determine the required intersection size. - Fill the rest of \(B\) from outside \(A\).

Solution

1. \(P(A)=\frac{6}{12}=\frac12\) and \(P(B)=\frac{4}{12}=\frac13\). 2. Independence requires \(P(A\cap B)=\frac16\). 3. The intersection must therefore contain \(12\cdot\frac16=2\) outcomes. Choose any \(2\) from \(A\) and \(2\) from \(A^c\).

Answer

\(A\cap B\) must contain \(2\) outcomes. Choose \(B\) with \(2\) outcomes from \(A\) and \(2\) from \(A^c\).
54738712
Assume \(P(A)>0\) and \(P(B)>0\). Prove that \(P(A\mid B)=P(A)\) if and only if \(P(B\mid A)=P(B)\). State how this establishes the symmetry of independence.

Hints

- Convert one conditional equality into a statement about the joint probability. - Use the positivity assumptions when dividing. - Repeat the algebra in reverse for the converse.

Solution

1. \(P(A\mid B)=P(A)\) implies \(\frac{P(A\cap B)}{P(B)}=P(A)\), so \(P(A\cap B)=P(A)P(B)\). 2. Dividing this equality by \(P(A)\) gives \(P(B\mid A)=P(B)\). 3. Reversing the same steps proves the converse. 4. Both conditional equalities are equivalent to the symmetric product criterion for independence.

Answer

The two conditional equalities are equivalent because each is equivalent to \(P(A\cap B)=P(A)P(B)\). Thus independence does not depend on which event is named first.
54739212
Events \(A,B,C\) satisfy \(P(A)=P(B)=P(C)=\frac12\), every pairwise intersection has probability \(\frac14\), and \(P(A\cap B\cap C)=\frac18\). Are the events mutually independent?

Hints

- Mutual independence requires more than pairwise checks. - Verify the probability of every nontrivial intersection. - Compare the triple intersection with the product of all three marginals.

Solution

1. The pairwise intersections equal the products of their marginals. 2. The triple intersection equals \((\frac12)^3=\frac18\). 3. These conditions establish mutual independence for the three events.

Answer

Yes. Both all pairwise product conditions and the triple-product condition hold.
54735212
The displayed event diagram represents two independent events \(A\) and \(B\), with \(P(A)=p\) and \(P(B)=q\). Find the probabilities of the four labeled regions. Use them to prove that each pair \((A,B)\), \((A,B^c)\), \((A^c,B)\), and \((A^c,B^c)\) is independent.
Figure for problem 547352

Hints

- Start with the one joint probability supplied by independence. - Obtain adjacent regions by subtracting from a margin. - Factor each result into the relevant marginal probabilities.

Solution

1. Independence gives \(P(A\cap B)=pq\). 2. \(P(A\cap B^c)=P(A)-P(A\cap B)=p-pq=p(1-q)\). 3. \(P(A^c\cap B)=q-pq=(1-p)q\). 4. The remaining region is \(1-p-q+pq=(1-p)(1-q)\). 5. Each joint-region probability equals the product of the corresponding marginal probabilities, proving all four independence statements.

Answer

The four joint probabilities are \(pq\), \(p(1-q)\), \((1-p)q\), and \((1-p)(1-q)\). Thus complementing either or both independent events preserves independence.
54735412
Events \(A\) and \(B\) are independent, with \(0<P(A)<1\) and \(P(B)>0\). Let \(C=A\cap B\). Can \(A\) and \(C\) be independent? Derive the condition and explain why it fails under the stated assumptions.

Hints

- Use the subset relationship between the original event and the new event. - Write the product criterion for the two events being tested. - Check whether the intersection event has positive probability before dividing.

Solution

1. Because \(C\subseteq A\), \(A\cap C=C\). 2. Independence of \(A\) and \(C\) would require \(P(C)=P(A)P(C)\). 3. Since \(P(C)=P(A)P(B)>0\), divide by \(P(C)\) to obtain \(P(A)=1\). 4. This contradicts \(0<P(A)<1\), so \(A\) and \(A\cap B\) are not independent.

Answer

No. Independence would force \(P(A)=1\), contradicting \(0<P(A)<1\). Although \(A\) is independent of \(B\), it is generally dependent on the smaller event \(A\cap B\).
54735512
The probability tree shows how a customer is assigned to one of two email systems and how two messages are processed through that same system. Let \(A\) be the event that the first message is flagged and \(B\) the event that the second message is flagged. a) Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\). b) Determine whether \(A\) and \(B\) are independent without conditioning on the system. c) Explain how the messages can be conditionally independent given the system but dependent in the combined population.
Figure for problem 547355

Hints

- Separate calculations within each system from calculations after the systems are mixed. - The same hidden system assignment affects both message outcomes. - Compare the joint probability with the product of the marginal probabilities.

Solution

1. Each message has marginal flag probability \(P(A)=P(B)=\frac12\cdot0.20+\frac12\cdot0.80=0.50\). 2. Conditional independence within each system gives \(P(A\cap B)=\frac12\cdot(0.20)^2+\frac12\cdot(0.80)^2=0.34\). 3. Marginal independence would require \(P(A\cap B)=P(A)P(B)=0.25\), but \(0.34\ne0.25\). 4. Given the system, the message outcomes are generated independently. Without that condition, both outcomes share the same unobserved system assignment, which creates positive dependence.

Answer

a) \(P(A)=P(B)=0.50\) and \(P(A\cap B)=0.34\). b) The events are not marginally independent because \(0.34\ne0.50\cdot0.50\). c) They are independent within each system, but mixing systems with different flag rates creates dependence through the shared system assignment.
54735912
Independent events \(A\) and \(B\) have the same nonzero probability \(p\). Given that at least one event occurred, the conditional probability that \(A\) occurred is \(\frac35\). Find \(p\).

Hints

- Express the probability of the union using the overlap required by independence. - The favorable event is contained in the conditioning event. - Simplify before solving for the common probability.

Solution

1. \(P(A\cap B)=p^2\), so \(P(A\cup B)=2p-p^2\). 2. Since \(A\subseteq A\cup B\), \(P(A\mid A\cup B)=\frac{p}{2p-p^2}=\frac{1}{2-p}\). 3. Solve \(\frac{1}{2-p}=\frac35\): \(5=6-3p\), so \(p=\frac13\).

Answer

\(p=\frac13\).
54736012
Events \(B_1,B_2,B_3\) form a partition, with each having positive probability. Suppose \(P(A\mid B_1)=P(A\mid B_2)=P(A\mid B_3)=0.40\). Prove that \(A\) is independent of every event formed by taking a union of some of the partition cells.

Hints

- Convert each conditional equality into a joint probability. - Use the fact that partition cells are disjoint. - Compare the resulting joint probability with the marginal probability of the union.

Solution

1. For each cell, \(P(A\cap B_i)=0.40P(B_i)\). 2. For a union \(U\) of selected partition cells, the corresponding intersections \(A\cap B_i\) are disjoint. 3. Therefore \(P(A\cap U)=\sum 0.40P(B_i)=0.40P(U)\). 4. The full partition also gives \(P(A)=\sum_i0.40P(B_i)=0.40\). 5. Hence \(P(A\cap U)=P(A)P(U)\), so \(A\) is independent of every such union.

Answer

Because the conditional probability of \(A\) is the same \(0.40\) in every partition cell, \(P(A)=0.40\) and \(P(A\cap U)=0.40P(U)\) for any union \(U\) of cells. Thus \(A\) and \(U\) are independent.
54736212
Two fair bits \(X\) and \(Y\) are generated independently. Define \(A=\{X=0\},\quad B=\{Y=0\},\quad C=\{X=Y\}\). Show that the events are pairwise independent but not mutually independent.

Hints

- List the four equally likely bit pairs. - Check every pairwise product condition. - Mutual independence also requires the triple-intersection condition.

Solution

1. Each event has probability \(\frac12\). 2. Every pairwise intersection has one of four equally likely outcomes, so each has probability \(\frac14=\frac12\cdot\frac12\). 3. \(A\cap B\cap C=\{(0, 0)\}\) has probability \(\frac14\), not \((\frac12)^3=\frac18\), so mutual independence fails.

Answer

The events are pairwise independent, but not mutually independent because the triple intersection has probability \(\frac14\) instead of \(\frac18\).
54736612
Events \(A\) and \(B\) are independent, with \(P(B)=\frac12\). Let \(D=A\triangle B\), the event that exactly one of \(A\) and \(B\) occurs. Prove that \(A\) and \(D\) are independent, regardless of \(P(A)\).

Hints

- Rewrite the symmetric difference as two disjoint pieces. - Use independence for an event paired with the complement of the other. - Compare the intersection with the product of marginals.

Solution

1. Let \(P(A)=p\). Independence gives \(P(A\cap B^c)=p\cdot\frac12=\frac p2\). 2. \(D=(A\cap B^c)\cup(A^c\cap B)\), so \(P(D)=\frac p2+\frac{1-p}{2}=\frac12\). 3. \(A\cap D=A\cap B^c\), with probability \(\frac p2\). 4. \(P(A)P(D)=p\cdot\frac12=\frac p2=P(A\cap D)\), so \(A\) and \(D\) are independent.

Answer

\(P(D)=\frac12\) and \(P(A\cap D)=\frac{P(A)}2=P(A)P(D)\). Therefore \(A\) and \(A\triangle B\) are independent.
54737812
Construct a probability-space example with three distinct events showing that independence is not transitive: \(A\) is independent of \(B\), and \(B\) is independent of \(C\), but \(A\) is not independent of \(C\).

Hints

- Use a product sample space so one event can depend only on the second coordinate. - Choose two distinct events depending on the first coordinate, with one contained in the other. - Verify each of the three pairwise product conditions separately.

Solution

1. Use the six equally likely ordered pairs \(S=\{1,2,3\}\times\{0,1\}\). Let \(A=\{X=1\}\), \(B=\{Y=0\}\), and \(C=\{X\in\{1,2\}\}\). 2. \(P(A)=\frac13\), \(P(B)=\frac12\), and \(P(A\cap B)=\frac16=P(A)P(B)\), so \(A\) and \(B\) are independent. 3. \(P(C)=\frac23\), and \(P(B\cap C)=\frac13=P(B)P(C)\), so \(B\) and \(C\) are independent. 4. Since \(A\subset C\), \(P(A\cap C)=\frac13\), while \(P(A)P(C)=\frac29\). Thus \(A\) and \(C\) are not independent.

Answer

For example, on \(S=\{1,2,3\}\times\{0,1\}\) with equally likely outcomes, take \(A=\{X=1\}\), \(B=\{Y=0\}\), and \(C=\{X\in\{1,2\}\}\). Then \(A\) is independent of \(B\), and \(B\) is independent of \(C\), but \(A\) is not independent of \(C\).
54738612
Prove that an event \(A\) is independent of every event in its probability space if and only if \(P(A)\) is \(0\) or \(1\).

Hints

- Test the “every event” claim using the event itself. - Solve the resulting equation for its probability. - Verify separately that each endpoint probability works with an arbitrary event.

Solution

1. If \(A\) is independent of every event, it is independent of itself. Thus \(P(A)=P(A)^2\), so \(P(A)\in\{0,1\}\). 2. If \(P(A)=0\), then \(P(A\cap B)=0=P(A)P(B)\) for every \(B\). 3. If \(P(A)=1\), then \(P(A^c)=0\), so \(P(A\cap B)=P(B)=P(A)P(B)\) for every \(B\). 4. Therefore the condition is both necessary and sufficient.

Answer

An event is independent of every event exactly when its probability is \(0\) or \(1\).
54738812
Events \(A,B,C\) are mutually independent. Prove that \(A\cup B\) is independent of \(C\), even when \(A\) and \(B\) overlap.

Hints

- Intersect the union with \(C\) and apply inclusion-exclusion. - Replace pair and triple intersections using mutual independence. - Factor the probability of \(C\) and recognize a union probability.

Solution

1. \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\). 2. Inclusion-exclusion gives \(P((A\cup B)\cap C)=P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)\). 3. Mutual independence changes this to \(P(A)P(C)+P(B)P(C)-P(A)P(B)P(C)\). 4. Factor \(P(C)\): the result is \(P(C)[P(A)+P(B)-P(A)P(B)]\). 5. Since mutual independence includes independence of \(A\) and \(B\), the bracket is \(P(A\cup B)\). Thus the product criterion holds.

Answer

\(P((A\cup B)\cap C)=P(A\cup B)P(C)\), so \(A\cup B\) and \(C\) are independent.
54738912
A categorical variable \(X\) has categories \(x_1,x_2,x_3\), and \(Y\) has categories \(y_1,y_2\). Their joint probabilities are <table> <tr><th></th><th>\(y_1\)</th><th>\(y_2\)</th><th>Total</th></tr> <tr><td>\(x_1\)</td><td>\(0.25\)</td><td>\(0.05\)</td><td>\(0.30\)</td></tr> <tr><td>\(x_2\)</td><td>\(0.05\)</td><td>\(0.25\)</td><td>\(0.30\)</td></tr> <tr><td>\(x_3\)</td><td>\(0.20\)</td><td>\(0.20\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> Show that \(X\) and \(Y\) are dependent. Then merge \(x_1\) and \(x_2\) into one category and show that the merged variable is independent of \(Y\). Explain what the example demonstrates.

Hints

- Test one original cell against the product of its margins. - Add corresponding cells when the two categories are merged. - Recheck the product criterion after aggregation and compare the conclusions.

Solution

1. For the original table, independence would require \(P(x_1,y_1)=P(x_1)P(y_1)=0.30\cdot0.50=0.15\), but the actual cell is \(0.25\). Thus \(X\) and \(Y\) are dependent. 2. After merging \(x_1,x_2\), the merged row is \((0.30, 0.30)\) with margin \(0.60\). The \(x_3\) row is \((0.20, 0.20)\) with margin \(0.40\). 3. Each merged cell equals row margin times the corresponding column margin: \(0.60\cdot0.50=0.30\) and \(0.40\cdot0.50=0.20\). 4. Therefore the merged variable and \(Y\) are independent. Aggregation can hide dependence present among finer categories.

Answer

The original variables are dependent because, for example, \(0.25\ne0.30\cdot0.50\). After merging \(x_1,x_2\), the table has rows \((0.30, 0.30)\) and \((0.20, 0.20)\), which are independent of \(Y\). Category merging can conceal dependence.
54739012
Two categorical variables \(X\) and \(Y\) are independent. Several categories of \(X\) are merged into a new category \(G\), while the categories of \(Y\) are unchanged. Prove that the recoded variable remains independent of \(Y\). State the key role of disjoint category events.

Hints

- Express the merged category as a union of original categories. - Use independence on each original category separately. - Add probabilities only after confirming the category pieces are disjoint.

Solution

1. Let \(G\) be the union of disjoint original X-category events \(X_1,\ldots,X_k\). 2. For any Y-category event \(H\), independence gives \(P(X_i\cap H)=P(X_i)P(H)\). 3. The intersections \(X_i\cap H\) are disjoint, so \(P(G\cap H)=\sum_iP(X_i)P(H)=P(H)\sum_iP(X_i)\). 4. Since \(\sum_iP(X_i)=P(G)\), \(P(G\cap H)=P(G)P(H)\). 5. Thus merging categories of one variable preserves independence.

Answer

The recoded category \(G\) satisfies \(P(G\cap H)=P(G)P(H)\) for every category \(H\) of \(Y\). Therefore the recoded variable remains independent of \(Y\).

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