Independent events and unions

Learning goals

Tests independence using conditional probabilities or product relationships.
Distinguishes mutually exclusive events from independent events.
Uses multiplication rules for independent events.
Interprets independence in context.
Computes union probabilities using the addition rule.

Difficulty: 1 – 5

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Independent events and unions 1

Events \(A\) and \(B\) are independent, with \(P(A)=0.50\) and \(P(B)=0.40\). Find \(P(A\cap B)\).

Hints

- Use the defining product relationship for independent events. - The question asks for both events to occur. - Combine the two marginal probabilities according to the independence condition.

Solution

1. Independent events satisfy \(P(A\cap B)=P(A)P(B)\). 2. Therefore \(P(A\cap B)=0.50\cdot0.40=0.20\).

Answer

\(0.20\)

Independent events and unions 2

Events \(A\) and \(B\) satisfy \(P(A)=0.35\), \(P(B)=0.25\), and \(P(A\cap B)=0.10\). Find \(P(A\cup B)\).

Hints

- The union includes outcomes in either event. - The intersection has been included once in each marginal probability. - Adjust the sum so the overlap is counted only once.

Solution

1. Add the two event probabilities and subtract the overlap once. 2. \(P(A\cup B)=0.35+0.25-0.10=0.50\).

Answer

\(0.50\)

Independent events and unions 3

A fair coin is flipped twice. The first flip is heads. Does knowing the first result change the probability of heads on the second flip? Explain.

Hints

- What makes a coin flip fair? - Does the first flip physically change the coin before the second flip? - What would have to happen to the second chance for the events to be dependent?

Solution

1. Each flip of a fair coin has a \(\frac{1}{2}\) chance of heads. 2. The second flip is not affected by the first result, so its chance of heads remains \(\frac{1}{2}\). 3. The results of the two flips are independent.

Answer

No. The probability of heads on the second flip stays \(\frac{1}{2}\). The two flips are independent.

Independent events and unions 4

Two events have \(P(A)=0.30\), \(P(B)=0.50\), and \(P(A\cap B)=0.15\). Are A and B independent? Check using the multiplication rule.

Hints

- Would knowing that \(A\) happened have to change the chance of \(B\)? - See whether the given overlap is consistent with the two separate probabilities. - Check that your conclusion follows from the numbers rather than the event names.

Solution

1. For independent events, the probability of both is the product of the separate probabilities. 2. Here \(P(A)P(B)=0.30\cdot0.50=0.15\), which matches \(P(A\cap B)\). 3. Therefore A and B are independent.

Answer

Yes. \(0.30\cdot0.50=0.15=P(A\cap B)\).

Independent events and unions 5

The table shows which of \(20\) museum visitors took a guided tour and visited the gift shop. <table> <tr><th></th><th>Gift shop</th><th>No gift shop</th></tr> <tr><td>Guided tour</td><td>\(5\)</td><td>\(7\)</td></tr> <tr><td>No guided tour</td><td>\(4\)</td><td>\(4\)</td></tr> </table> What is the probability that a randomly selected visitor took a tour **or** visited the shop (or both)?

Hints

- Which cell contains visitors who did neither activity? - Would someone who toured the museum and shopped belong to the requested event? - How might the group who did neither help you check your result?

Solution

1. Only the \(4\) visitors in the lower-right cell did neither activity. 2. The other \(20-4=16\) visitors did at least one. 3. Thus the probability is \(\frac{16}{20}=\frac{4}{5}\).

Answer

\(\frac{4}{5}\), or \(0.80\).

Independent events and unions 6

One ball is drawn at random from jar a), then one from jar b). The first ball from jar a) is green. Does that result change the probability that the ball from jar b) is red? Give the probability and explain.

Hints

- Which jar will supply the second ball? - Did removing a ball from jar a) change the contents of jar b)? - Find the red share in jar b) before and after hearing the first result.

Solution

1. Jar b) has \(3\) red balls and \(1\) blue ball, so the chance of red from jar b) is \(\frac{3}{4}\). 2. The first ball came from a different jar, so nothing in jar b) changed. 3. The chance remains \(\frac{3}{4}\); the two draws are independent.

Answer

No. The chance of red from jar b) stays \(\frac{3}{4}\), because the first draw does not change jar b).

Independent events and unions 7

A \(3\times2\) joint relative-frequency table has cells \( \begin{pmatrix} 0.12&0.18\\ 0.20&0.10\\ 0.08&0.32 \end{pmatrix}. \) Find the relative frequency of the compound event “row 1 or column 1.” Show which joint cells belong to the event and avoid double-counting their overlap.

Hints

- Mark the cells satisfying each part of the “or” statement. - Identify any cell selected twice. - Sum the distinct joint cells exactly once.

Solution

1. Row 1 contains \(0.12+0.18=0.30\). 2. Column 1 contains \(0.12+0.20+0.08=0.40\). 3. The upper-left cell \(0.12\) lies in both descriptions. 4. The union frequency is \(0.30+0.40-0.12=0.58\), equivalently the sum of cells \(0.12+0.18+0.20+0.08\).

Answer

\(0.58\).

Independent events and unions 8

A fair die is rolled many times. Ece 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 \(\frac{1}{2}\).

Independent events and unions 9

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 \(0.10\).

Independent events and unions 10

In a school survey, \(30\%\) of all students ride the bus, and \(40\%\) participate in a sports club. Suppose bus use and sports-club membership are independent. a) What percentage of sports-club members ride the bus? b) What percentage of all students both ride the bus and participate in a sports club? c) Why is the answer to part a) not \(12\%\)?

Hints

- Under independence, does learning about sports-club membership change the bus-riding chance? - What event must happen for a student to belong to both groups? - Which denominator is used in each percentage?

Solution

1. Independence means the bus-riding chance is unchanged inside the sports-club group, so part a) is \(30\%\). 2. The share who meet both conditions is \(0.30\cdot0.40=0.12=12\%\). 3. The \(12\%\) is a share of everyone surveyed, while the \(30\%\) is a share within sports-club members.

Answer

a) \(30\%\). b) \(12\%\). c) Part a) uses only sports-club members as the reference group; \(12\%\) describes the whole survey.

Independent events and unions 11

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.

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 \(\frac13\).

Independent events and unions 12

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?

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 \(0.40\).

Independent events and unions 13

Thiago 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.

Independent events and unions 14

A point \((X, Y)\) is chosen uniformly from the displayed unit square. Event \(A\) is the region to the left of the vertical dashed boundary, and event \(B\) is the region above the horizontal dashed boundary. Read the boundary values from the axes and use areas to show that \(A\) and \(B\) are independent.

Hints

- Read each dashed boundary from the corresponding axis before finding any area. - For a uniformly chosen point, an event's probability equals its fraction of the unit-square area. - Compare the overlap area with the product of the two event probabilities.

Solution

1. The vertical boundary is at \(x=0.30\), so \(P(A)=0.30\). 2. The horizontal boundary is at \(y=0.60\), so \(P(B)=1-0.60=0.40\). 3. The intersection is the upper-left rectangle, with area \(0.30\cdot0.40=0.12\). 4. Since \(P(A\cap B)=0.12=P(A)P(B)\), the events are independent.

Answer

\(P(A)=0.30\), \(P(B)=0.40\), and \(P(A\cap B)=0.12=0.30\cdot0.40\). Therefore \(A\) and \(B\) are independent.

Independent events and unions 15

This table shows the results of \(100\) observations. <table> <tr><th></th><th>\(B\)</th><th>not \(B\)</th></tr> <tr><td>\(A\)</td><td>\(24\)</td><td>\(16\)</td></tr> <tr><td>not \(A\)</td><td>\(36\)</td><td>\(24\)</td></tr> </table> Does knowing whether \(A\) occurred change the observed chance of \(B\)? Use conditional relative frequencies to check.

Hints

- Compare the part of each row that falls in column \(B\). - Make sure each percentage uses the total for its own row. - If the two percentages match, what does that tell you about knowing \(A\)?

Solution

1. Among the \(40\) observations in \(A\), the relative frequency of \(B\) is \(\frac{24}{40}=0.60\). 2. Among the \(60\) observations outside \(A\), the relative frequency of \(B\) is \(\frac{36}{60}=0.60\). 3. The rates match, so in this table knowing whether \(A\) occurred does not change the observed chance of \(B\).

Answer

No. The two conditional relative frequencies are both \(0.60\). The observed table satisfies the independence condition.

Independent events and unions 16

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 \(0.08\).

Independent events and unions 17

Events \(A\) and \(B\) are independent, with \(P(A)=0.60\) and \(P(B)=0.50\). Find \(P(A\cup B)\).

Hints

- The union formula needs the intersection probability. - Use independence to determine the overlap before computing the union. - Count outcomes in both events only once.

Solution

1. Independence gives \(P(A\cap B)=0.60\cdot0.50=0.30\). 2. Apply the union formula: \(P(A\cup B)=0.60+0.50-0.30=0.80\).

Answer

\(0.80\)

Independent events and unions 18

The displayed tree shows three independent attempts of the same process. S means success and F means failure. Find the probability of at least one success in the three attempts.

Hints

- Read the success and failure probabilities from one stage of the tree. - Consider the complement of the event “at least one success.” - Independence lets you combine the probabilities along the all-failure path.

Solution

1. From the tree, each attempt has success probability \(0.70\) and failure probability \(0.30\). 2. The complement of “at least one success” is “all three attempts fail.” 3. By independence, \(P(FFF)=0.30^3=0.027\). 4. Therefore the probability of at least one success is \(1-0.027=0.973\).

Answer

\(0.973\)

Independent events and unions 19

Two pairs of events have the same marginal probabilities: \(P(A)=P(C)=0.20\) and \(P(B)=P(D)=0.30\). Events \(A\) and \(B\) are mutually exclusive, while events \(C\) and \(D\) are independent. a) Find \(P(A\cup B)\). b) Find \(P(C\cup D)\). c) Explain why the two union probabilities are different even though the marginal probabilities match.

Hints

- The two relationships impose different conditions on the intersection. - Determine the overlap for each pair before using the union formula. - Matching marginal probabilities do not force matching union probabilities.

Solution

1. Since \(A\) and \(B\) are mutually exclusive, \(P(A\cap B)=0\), so \(P(A\cup B)=0.20+0.30=0.50\). 2. Since \(C\) and \(D\) are independent, \(P(C\cap D)=0.20\cdot0.30=0.06\). 3. Therefore \(P(C\cup D)=0.20+0.30-0.06=0.44\). 4. The mutually exclusive pair has no overlap, while the independent pair has positive overlap, so their unions differ.

Answer

a) \(0.50\) b) \(0.44\) c) The mutually exclusive events have intersection probability \(0\), while the independent events have intersection probability \(0.06\).

Independent events and unions 20

The numbered dots are equally likely outcomes. Can events A and B happen together? Are they independent? Explain both answers using probabilities.

Hints

- Are any numbered dots inside both circles? - Count how many dots are in A and in B separately. - Compare the chance of both events with the product of their individual chances.

Solution

1. A contains two of the six outcomes, and B contains two of the six outcomes. They share no outcome. 2. Therefore A and B are mutually exclusive, with \(P(A\cap B)=0\). 3. Since \(P(A)P(B)=\frac{2}{6}\cdot\frac{2}{6}=\frac{1}{9}\ne0\), the events are not independent.

Answer

They are mutually exclusive because they never occur together. They are not independent because \(P(A\cap B)=0\) but \(P(A)P(B)=\frac{1}{9}\).

Independent events and unions 21

A warehouse recorded \(100\) deliveries. Of these, \(60\) used refrigerated trucks, and \(24\) were both refrigerated and on time. In this record, refrigeration and on-time arrival are independent. Complete the two-way table. <table> <tr><th></th><th>On time</th><th>Late</th><th>Total</th></tr> <tr><td>Refrigerated</td><td>\(24\)</td><td>\(\square\)</td><td>\(60\)</td></tr> <tr><td>Not refrigerated</td><td>\(\square\)</td><td>\(\square\)</td><td>\(\square\)</td></tr> <tr><td>Total</td><td>\(\square\)</td><td>\(\square\)</td><td>\(100\)</td></tr> </table>

Hints

- Does the on-time share differ between refrigerated and other deliveries if the events are independent? - What could the given refrigerated counts tell you about all deliveries? - Check that each row and column fits the recorded total.

Solution

1. Among refrigerated deliveries, \(\frac{24}{60}=0.40\) arrived on time. Independence means the same \(0.40\) applies to all \(100\) deliveries. 2. Thus the on-time total is \(40\), and the late total is \(60\). 3. Refrigerated/late is \(60-24=36\); not refrigerated/on time is \(40-24=16\); not refrigerated/late is \(60-36=24\). 4. The not-refrigerated row total is \(100-60=40\).

Answer

<table> <tr><th></th><th>On time</th><th>Late</th><th>Total</th></tr> <tr><td>Refrigerated</td><td>\(24\)</td><td>\(36\)</td><td>\(60\)</td></tr> <tr><td>Not refrigerated</td><td>\(16\)</td><td>\(24\)</td><td>\(40\)</td></tr> <tr><td>Total</td><td>\(40\)</td><td>\(60\)</td><td>\(100\)</td></tr> </table>

Independent events and unions 22

A bookstore puts a bookmark in \(60\%\) of its orders. Separately, it randomly includes a coupon in \(75\%\) of orders. The two choices are independent. What is the probability that an order contains a bookmark **but no coupon**?

Hints

- What fraction of orders do not receive a coupon? - Does knowing about the bookmark change that fraction? - Consider the two conditions together, not either one separately.

Solution

1. The chance of no coupon is \(1-0.75=0.25\). 2. Coupon selection and bookmark selection are independent, as are having a bookmark and not having a coupon. 3. Multiply: \(0.60\cdot0.25=0.15\).

Answer

\(0.15\), or \(15\%\).

Independent events and unions 23

Events A and B are independent. The probability of A is \(0.60\), and the probability of both A and B is \(0.18\). Find \(P(B)\).

Hints

- Would the overlap be determined by a different rule if the events depended on each other? - Compare the size of the overlap with the known chance of \(A\). - Check that the missing chance stays between \(0\) and \(1\).

Solution

1. Independence gives \(P(A\cap B)=P(A)P(B)\). 2. Substitute the known values: \(0.18=0.60P(B)\). 3. Divide to find \(P(B)=\frac{0.18}{0.60}=0.30\).

Answer

\(P(B)=0.30\).

Independent events and unions 24

A basketball player makes each free throw with probability \(0.20\), independently of earlier shots. The player has missed three times in a row. a) What is the probability of making the next shot? b) What is the probability of making at least one of the next two shots? c) Explain why the three earlier misses do not change either answer.

Hints

- Decide whether the player's success chance changes after a miss. - For two upcoming attempts, think about what it means not to succeed on either. - Keep the earlier shots separate from the two future shots.

Solution

1. Independence leaves the next-shot success probability at \(0.20\). 2. Missing both of the next two shots has probability \((0.80)^2=0.64\). 3. The probability of making at least one is \(1-0.64=0.36\). 4. Earlier results do not change the success probability of an independent future shot.

Answer

a) \(0.20\). b) \(0.36\). c) The shots are independent; previous misses do not alter the chances on future shots.

Independent events and unions 25

A security system presents a program with six independent challenges. The program answers each challenge correctly with probability \(p\). Access is granted only if all six challenges are answered correctly. Find the value of \(p\) that makes the overall probability of gaining access exactly \(50\%\).

Hints

- Write the overall probability as a product of six equal probabilities. - Form an equation involving \(p\). - Use a sixth root to reverse the sixth power.

Solution

1. Because all six independent challenges must be answered correctly, the overall probability is \(p^6\). 2. Set \(p^6=0.50\). 3. Take the sixth root: \(p=\sqrt[6]{0.50}\). 4. Numerically, \(p\approx 0.8909\).

Answer

\(p=\sqrt[6]{0.50}\approx 0.8909\), or about \(89.09\%\)

Independent events and unions 26

A report says: “Of \(90\) surveyed households, \(52\) have a pet, \(41\) have a garden, and \(70\) have a pet or a garden. The count \(70\) belongs in the ‘pet and garden’ cell.” Explain the error, find the correct “pet and garden” count, and complete the \(2\times2\) table.

Hints

- Distinguish the union from the intersection. - Use the overlap needed to avoid double-counting the two group totals. - The neither cell comes from the complement of the union.

Solution

1. The value \(70\) is a union count, not an intersection count. 2. Inclusion-exclusion gives the both count \(52+41-70=23\). 3. Pet only is \(29\), garden only is \(18\), and neither is \(90-70=20\).

Answer

<table> <tr><th></th><th>garden</th><th>no garden</th><th>Total</th></tr> <tr><td>pet</td><td>\(23\)</td><td>\(29\)</td><td>\(52\)</td></tr> <tr><td>no pet</td><td>\(18\)</td><td>\(20\)</td><td>\(38\)</td></tr> <tr><td>Total</td><td>\(41\)</td><td>\(49\)</td><td>\(90\)</td></tr> </table> The error is treating “or” as “and.”

Independent events and unions 27

A survey began with \(240\) people. Twelve are missing the row variable, \(17\) are missing the column variable, and \(5\) are missing both variables. How many people can be included in a complete-case two-way table? Explain why simply subtracting \(12+17\) is incorrect.

Hints

- Think of the two missingness groups as overlapping sets. - Count the overlap only once. - The table includes only records with both categorical values observed.

Solution

1. The number missing at least one variable is \(12+17-5=24\), because the \(5\) missing both were counted twice. 2. The complete-case count is \(240-24=216\). 3. Subtracting \(12+17\) would remove the five doubly missing records twice.

Answer

The complete-case table contains \(216\) people. Subtracting \(12+17\) is wrong because the \(5\) people missing both variables would be subtracted twice.

Independent events and unions 28

Two cameras inspect welds on bicycle frames. In a test, camera 2 found a flaw in \(82\%\) of cases when camera 1 found it, but in only \(31\%\) when camera 1 missed it. A simulation treats the cameras as independent. Do the test results support that choice? Name one feature of a weld that might explain the difference.

Hints

- What would independence say about camera 2's chance after learning camera 1's result? - Compare the two given percentages before deciding. - Could one feature of the weld affect both cameras?

Solution

1. Independence would mean learning camera 1's result does not change the chance that camera 2 detects the flaw. 2. The two recorded conditional rates, \(82\%\) and \(31\%\), are different, so the data do not support independence. 3. Large, easy-to-see flaws could be more likely to be found by both cameras.

Answer

No. Camera 2's detection rate differs with camera 1's result (\(82\%\) versus \(31\%\)). Flaw size could affect both cameras.

Independent events and unions 29

Of \(80\) robotics-club members, \(60\) also take part in band, \(52\) volunteer at school events, and \(40\) do both band and volunteering. a) What fraction of robotics-club members do **band or volunteering or both**? b) A student adds \(\frac{60}{80}+\frac{52}{80}\) and gets more than \(1\). Explain the mistake and show how to fix it.

Hints

- Which robotics members are included in both counts? - Is every student in that overlap being counted more than once in the sum? - Use the robotics group as the denominator throughout.

Solution

1. Adding the two group counts counts the \(40\) students in both groups twice. 2. The number in at least one group is \(60+52-40=72\), so the conditional share among robotics members is \(\frac{72}{80}=0.90\). 3. Subtract the shared \(\frac{40}{80}\) from the sum of the two conditional shares to avoid double-counting.

Answer

a) \(\frac{72}{80}=0.90\), or \(90\%\). b) The \(40\) students in both groups were counted twice. The corrected share is \(\frac{60+52-40}{80}=0.90\).

Independent events and unions 30

Six distinct books are arranged uniformly at random on a shelf. Let \(E\) be the event “book \(A\) appears before book \(B\),” and let \(F\) be the event “book \(C\) appears before book \(D\).” a) Find \(P(F)\). b) Find \(P(F\mid E)\) without listing all \(6!\) arrangements. c) Use your two results to determine whether \(E\) and \(F\) are independent.

Hints

- Compare the probability of the \(C,D\) order before and after restricting to arrangements with \(A\) before \(B\). - Look for a reversible change to an arrangement that switches the order of \(C\) and \(D\) without changing the order of \(A\) and \(B\). - Independence can be checked by comparing a marginal probability with the corresponding conditional probability.

Solution

1. In all shelf arrangements, swapping books \(C\) and \(D\) pairs every arrangement with \(C\) before \(D\) with one having \(D\) before \(C\). Thus \(P(F)=\frac12\). 2. Restrict to arrangements in \(E\), where \(A\) is before \(B\). Swapping \(C\) and \(D\) keeps the arrangement in \(E\), so exactly half of the arrangements in \(E\) satisfy \(F\). Hence \(P(F\mid E)=\frac12\). 3. Since \(P(F\mid E)=P(F)\) and \(P(E)>0\), the events \(E\) and \(F\) are independent.

Answer

a) \(P(F)=\frac12\) b) \(P(F\mid E)=\frac12\) c) Yes. The events are independent because conditioning on \(E\) does not change the probability of \(F\).

Independent events and unions 31

Two fair coins are flipped independently. The four equally likely outcomes are \(HH,HT,TH,TT\), with the first letter showing the first coin. a) What is the chance that both flips are heads? b) A friend tells you that **at least one flip was heads**. Under that condition, what is the chance the first flip was heads? c) Why can the answer to part b) differ from the usual \(\frac{1}{2}\) chance for the first coin, even though the two flips are independent?

Hints

- Start with all equally likely ordered outcomes. - Which outcome disappears when you learn there was at least one head? - Is knowing that one of the two was heads the same as knowing a particular coin's result?

Solution

1. Only \(HH\) has two heads, so the unconditional probability is \(\frac{1}{4}\). 2. The condition rules out \(TT\). Among \(HH,HT,TH\), the first flip is heads in \(HH\) and \(HT\), giving \(\frac{2}{3}\). 3. Independence says the result of one coin does not change the chance of the other. Learning that at least one head occurred is a different condition: it removes one of the four outcomes and changes the sample space.

Answer

a) \(\frac{1}{4}\). b) \(\frac{2}{3}\). c) The condition “at least one head” rules out \(TT\); it is not the same as learning the result of just one flip.

Independent events and unions 32

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.

Independent events and unions 33

In a sample of \(100\) observations, event \(A\) occurs \(30\) times, event \(B\) occurs \(50\) times, and both occur \(15\) times. a) Do the sample relative frequencies satisfy the product criterion for independence exactly? b) What may be concluded about this sample table? c) Why does this not prove that the population events are independent?

Hints

- Distinguish sample relative frequencies from population probabilities. - Apply the product criterion to the observed table first. - Consider whether one sample can establish an exact population property.

Solution

1. The sample margins are \(0.30\) and \(0.50\), and the sample joint frequency is \(0.15\). 2. Since \(0.15=0.30\cdot0.50\), the empirical table factors exactly and shows no sample association by this criterion. 3. Population independence is a statement about the underlying probability distribution. A finite sample can match the product relationship because of sampling variation, so inference would require a sampling model and an assessment of uncertainty.

Answer

a) Yes; \(0.15=0.30\cdot0.50\). b) The sample table is exactly independent descriptively. c) This does not prove population independence because the table is one finite random sample.

Independent events and unions 34

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 \(p=0.60\).

Independent events and unions 35

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 \(0.30\).

Independent events and unions 36

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

- Which flips does each event depend on? - Count the possibilities for the first two flips and the last three flips separately. - How do probabilities combine when the two blocks of trials do not affect one another?

Solution

1. Event \(A\) uses only the first two flips, while event \(B\) uses only the last three, so the two groups of flips do not affect each other. 2. For the first two flips, the matching results are HH and TT, so \(P(A)=\frac{2}{4}=\frac{1}{2}\). 3. For the last three flips, the one-head results are HTT, THT, and TTH, so \(P(B)=\frac{3}{8}\). 4. Combining one successful first block and one successful last block gives \(P(A\cap B)=\frac{1}{2}\cdot\frac{3}{8}=\frac{3}{16}\). 5. Since \(P(A\cap B)=P(A)P(B)\), the events are independent.

Answer

\(P(A)=\frac{1}{2}\), \(P(B)=\frac{3}{8}\), and \(P(A\cap B)=\frac{3}{16}\). The events are independent because they depend on separate groups of independent flips.

Independent events and unions 37

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.

Independent events and unions 38

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.

Hints

- Translate the operating rule of each diagram into an event involving the two components. - For the parallel system, consider the one way the whole system can fail. - Compare the two final reliability probabilities using a difference.

Solution

1. System a) is a series system, so both components must work. Its reliability is \(0.90^2=0.81\). 2. System b) fails only if both parallel components fail. That probability is \(0.10^2=0.01\). 3. Therefore system b) has reliability \(1-0.01=0.99\). 4. The difference is \(0.99-0.81=0.18\), so system b) is more reliable.

Answer

a) \(0.81\) b) \(0.99\) c) System b) is \(0.18\) more reliable.

Independent events and unions 39

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 \(\frac23\).

Independent events and unions 40

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: \(0.07\)

Independent events and unions 41

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 \(1\).

Independent events and unions 42

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 \(P(A)P(B)\).

Independent events and unions 43

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 \(P(A\cap B)=\frac14=P(A)P(B)\).

Independent events and unions 44

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 \(A^c\).

Independent events and unions 45

Let \(A\) be an event with probability \(p\). A student claims that \(A\) is independent of **itself**. a) What equation must \(p\) satisfy for the student's claim to be true? b) Find all possible values of \(p\). c) Explain in words why an event with a chance strictly between \(0\) and \(1\) cannot be independent of itself.

Hints

- What does “A and A both happen” mean? - What does the product rule for independent events require here? - Would learning that an event occurred change its chance if it was initially uncertain?

Solution

1. The intersection of \(A\) with itself is just \(A\), so independence would require \(p=p\cdot p\). 2. Solving \(p(1-p)=0\) gives \(p=0\) or \(p=1\). 3. Learning that an uncertain event happened would make its conditional chance \(1\), rather than its original chance \(p\); it therefore changes the chance.

Answer

a) \(p=p^2\). b) \(p=0\) or \(p=1\). c) Knowing an uncertain event happened changes its chance to certainty, so it cannot be independent of itself.

Independent events and unions 46

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.

Independent events and unions 47

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.

Independent events and unions 48

Events \(A\) and \(B\) are independent. Suppose \(P(A)=0.40\) and \(P(A\cup B)=0.70\). Find \(P(B)\).

Hints

- Introduce an unknown for the missing marginal probability. - Independence determines the intersection in terms of that unknown. - Substitute that expression into the union relationship and solve.

Solution

1. Let \(P(B)=p\). Independence gives \(P(A\cap B)=0.40p\). 2. Use the union formula: \(0.70=0.40+p-0.40p\). 3. Simplify to \(0.30=0.60p\), so \(p=0.50\).

Answer

\(0.50\)

Independent events and unions 49

A bicycle frame is checked against three safety standards. In a model, the chances a frame passes standards A, B, and C are \(0.90\), \(0.85\), and \(0.80\), respectively. No information is given about how passing one standard relates to passing the others. Find the smallest and largest possible chances that a frame passes all three. Explain how each extreme could happen.

Hints

- What is the most an intersection can contain compared with its smallest event? - Could the failure groups overlap or be completely separate? - Which arrangements of the failure groups make the union largest or smallest?

Solution

1. Passing all three requires passing C, so the chance cannot exceed \(0.80\). 2. Failure chances are \(0.10\), \(0.15\), and \(0.20\). Together, their union has chance at most \(0.45\), so passing all three has chance at least \(1-0.45=0.55\). 3. The lower limit is possible if the three failure groups are disjoint, making their union \(0.45\). 4. The upper limit is possible if every failure of A also fails B and every failure of B also fails C. Then all failures are contained in C's \(0.20\) failure group.

Answer

The chance of passing all three can range from \(0.55\) to \(0.80\). Disjoint failures attain \(0.55\); nested failures attain \(0.80\).

Independent events and unions 50

The displayed event diagram represents two independent events \(A\) and \(B\), with \(P(A)=p\) and \(P(B)=q\). The rectangle is the sample space. 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.

Hints

- Start with the joint probability supplied by independence. - Use the margins to determine the two one-event-only regions. - The fourth region is outside both events; compare each region with the product of its two relevant marginals.

Solution

1. Independence gives \(P(A\cap B)=pq\). 2. The A-only region is \(P(A\cap B^c)=P(A)-P(A\cap B)=p-pq=p(1-q)\). 3. The B-only region is \(P(A^c\cap B)=q-pq=(1-p)q\). 4. The neither region is \(1-p-q+pq=(1-p)(1-q)\). 5. Each region probability equals the product of the corresponding marginal probabilities: \(pq\), \(p(1-q)\), \((1-p)q\), and \((1-p)(1-q)\). Therefore complementing either or both events preserves independence.

Answer

Both: \(pq\) A only: \(p(1-q)\) B only: \((1-p)q\) Neither: \((1-p)(1-q)\) Each is the product of the appropriate marginal probabilities, so all four stated pairs are independent.

Independent events and unions 51

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 \(A\cap B\).

Independent events and unions 52

An email service randomly selects a filter and sends two messages through that **same** filter. Once a filter is selected, the first message's result does not affect the second message's chance of being flagged. Let \(A\) mean the first message is flagged and \(B\) mean the second is flagged. a) Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\). b) Are the two flag events independent if you do not know which filter was used? Check using the product rule. c) Why does using the same filter for both messages matter?

Hints

- Split the experiment by the two possible filters. - How is the chance of two flags calculated within a single selected filter? - Compare the overall both-flag chance with the product of the two overall single-flag chances.

Solution

1. The probability of a flag on either message is \(\frac{1}{2}\cdot0.20+\frac{1}{2}\cdot0.80=0.50\). 2. Both messages are flagged with probability \(\frac{1}{2}\cdot(0.20)^2+\frac{1}{2}\cdot(0.80)^2=0.34\). 3. Independence would require \(P(A\cap B)=0.50\cdot0.50=0.25\), but \(0.34\ne0.25\). 4. Both flag events depend on the same selected filter, so a high-flag filter makes both more likely even though messages are handled separately within that filter.

Answer

a) \(P(A)=P(B)=0.50\); \(P(A\cap B)=0.34\). b) No; \(0.34\ne0.25\). c) One filter choice affects the chances for both messages, creating a link between the results.

Independent events and unions 53

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

\(p=\frac13\).

Independent events and unions 54

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

- Before considering an arbitrary union of cells, determine what the three equal conditional probabilities force about the overall probability of \(A\). - Pick a union \(U\) of some partition cells. How can the partition structure help you express probabilities involving \(U\)? - What equality involving \(P(A\cap U)\), \(P(A)\), and \(P(U)\) would be enough to finish the proof?

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.

Independent events and unions 55

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 \(\frac18\).

Independent events and unions 56

A student enters a raffle with a chance \(p\) of winning, where \(0<p<1\), and also flips a fair coin. The raffle and coin flip are independent. Let \(A\) mean winning the raffle. Let \(D\) mean that **exactly one** of these two events happens: the student wins the raffle, or the coin shows heads. a) Find \(P(D)\) in terms of \(p\). b) Find \(P(A\cap D)\). c) Are \(A\) and \(D\) independent? Show this with the product rule.

Hints

- Which two different win/heads outcomes count as exactly one happening? - Does either of those outcomes allow a raffle win and a head together? - To check independence, which three probabilities must be compared?

Solution

1. The student wins and flips tails with probability \(p\cdot\frac{1}{2}\). The student loses and flips heads with probability \((1-p)\cdot\frac{1}{2}\). 2. These two outcomes cannot both happen, so \(P(D)=\frac{p}{2}+\frac{1-p}{2}=\frac{1}{2}\). 3. Event \(A\cap D\) means a raffle win and tails, so \(P(A\cap D)=\frac{p}{2}\). 4. Since \(P(A)P(D)=p\cdot\frac{1}{2}=\frac{p}{2}\), the events are independent.

Answer

a) \(\frac{1}{2}\). b) \(\frac{p}{2}\). c) Yes; \(P(A\cap D)=P(A)P(D)=\frac{p}{2}\).

Independent events and unions 57

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

- A small finite sample space makes it easier to verify each required relationship exactly. - Think about how one event could be independent of two different events for the same structural reason. - Your construction must make the product criterion succeed for two pairs and fail for the third pair.

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 \(C\).

Independent events and unions 58

Events \(A,B,C\) are mutually independent. Prove that \(A\cup B\) is independent of \(C\), even when \(A\) and \(B\) overlap.

Hints

- Decide which probability equality would establish the required independence. - The overlap between \(A\) and \(B\) must be accounted for rather than ignored. - Identify which consequences of mutual independence connect the relevant intersections to products of probabilities.

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 gives \(P(A\cap C)=P(A)P(C)\), \(P(B\cap C)=P(B)P(C)\), and \(P(A\cap B\cap C)=P(A)P(B)P(C)\). 4. Factoring \(P(C)\) gives \(P((A\cup B)\cap C)=P(C)[P(A)+P(B)-P(A)P(B)]\). 5. Since \(A\) and \(B\) are independent, \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\). Therefore \(P((A\cup B)\cap C)=P(A\cup B)P(C)\), so \(A\cup B\) and \(C\) are independent.

Answer

\(P((A\cup B)\cap C)=P(A\cup B)P(C)\), so \(A\cup B\) and \(C\) are independent.

Independent events and unions 59

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.

Independent events and unions 60

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

- Focus on one merged category \(G\) and one unchanged category \(H\) of \(Y\). What must be shown about that pair? - Think about how \(G\) is built from the original categories of \(X\), and what independence already tells you before the recoding. - Which property of distinct categories of the same categorical variable allows their probabilities to be combined without double-counting?

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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Statistics math worksheets

Independent events and unions

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