Free math worksheets

- List all possible die outcomes. - Remove exactly the outcomes that belong to \(A\).

1. The event is \(A=\{2,4,6\}\). 2. The remaining outcomes in \(S=\{1,2,3,4,5,6\}\) form the complement. 3. Therefore, \(A^c=\{1,3,5\}\).

\(A^c=\{1,3,5\}\).

- What is the total probability of all possible outcomes? - The complement includes every outcome that is not in \(A\).

1. The probabilities of an event and its complement sum to \(1\). 2. Therefore, \(P(A^c)=1-0.37=0.63\).

\(P(A^c)=0.63=63\%\).

- When is the statement “at least one head” not true? - Consider the two tosses together.

1. The complement of “at least one head” is “no heads.” 2. With two tosses, no heads means two tails. 3. Therefore, \(A^c=\{TT\}\).

The complement is “both tosses are tails,” and \(A^c=\{TT\}\).

- Negate the defining property without listing every outcome. - Keep the original range from \(1\) through \(50\).

1. The complement contains every number in the sample space that does not belong to \(A\). 2. Therefore, \(A^c\) is the event that the selected number is not divisible by \(5\).

\(A^c\): “The selected number is not a multiple of \(5\).”

- Identify the two statements as an event and its complement. - Compare the given probability with the full \(100\%\).

1. “It does not rain” is the complement of “it rains.” 2. Subtract from the total probability: \(100\%-28\%=72\%\).

The probability that it does not rain during the trip is \(72\%\).

- Event \(A\) and its complement cover all possible outcomes. - Check that the two probabilities add to \(1\).

1. An event and its complement have probabilities that sum to \(1\). 2. Therefore, \(P(A)=1-0.62=0.38\).

\(P(A)=0.38=38\%\).

- Identify which possible colors are already included in event \(A\). - Combine all remaining colors into one statement.

1. Event \(A\) contains all red and blue cards. 2. The complement contains the remaining colors, green and yellow. 3. Therefore, \(A^c\) is the event “the card is green or yellow.”

\(A^c\): “The card is green or yellow.”

- Write \(1\) with the same denominator as the given probability. - Check whether the resulting fraction can be reduced.

1. Use the complement rule: \(P(A^c)=1-\frac{7}{20}\). 2. Write \(1\) with denominator \(20\): \(P(A^c)=\frac{20}{20}-\frac{7}{20}=\frac{13}{20}\). 3. The fraction \(\frac{13}{20}\) is already in lowest terms.

\(P(A^c)=\frac{13}{20}\).

- Use the fact that there is no third response option. - Find the missing part of the full \(100\%\).

1. Because there are only two response options, \(A^c\) is the event “the response is No.” 2. Use the complement rule: \(P(A^c)=100\%-84\%=16\%\).

\(A^c\) is “the response is No,” and \(P(A^c)=16\%\).

- List the possible numbers of red or blue pens: \(0\), \(1\), \(2\), or \(3\). - Remove the cases included in the stated event. The cases left over form the complement. - “At least” and “at most” often lead to complementary descriptions, but check the boundary value carefully. - Can you describe the same outcome by referring to the other color?

1. Event \(A\) is the case of three blue pens. Every other possible color count includes at least one red pen. Therefore, \(A^c\) is: At least one pen is red. 2. Event \(B\) includes two or three red pens. Its complement includes zero or one red pen. Therefore, \(B^c\) is: At most one pen is red. 3. Event \(C\) includes zero or one blue pen. Its complement includes two or three blue pens. Therefore, \(C^c\) is: At least two pens are blue. 4. Event \(D\) includes exactly two red pens. Its complement includes every other possible number of red pens. Therefore, \(D^c\) is: Zero, one, or three pens are red.

a) \(A^c\): At least one pen is red. b) \(B^c\): At most one pen is red. c) \(C^c\): At least two pens are blue. d) \(D^c\): Zero, one, or three pens are red.

- List the integers from \(1\) through \(20\). - Mark the outcomes in each event. The unmarked outcomes belong to the complement. - What inequality represents “at most”? - Remember that \(1\) is not a prime number.

1. The complements are: \(E_1^c\), the number is not prime; \(E_2^c\), the number is not a multiple of \(4\); \(E_3^c\), the number is at most \(15\); and \(E_4^c\), the number is greater than \(18\). 2. The prime numbers from \(1\) through \(20\) are \(2,3,5,7,11,13,17,19\), so \(E_1^c\) has \(20-8=12\) outcomes. The multiples of \(4\) are \(4,8,12,16,20\), so \(E_2^c\) has \(20-5=15\) outcomes. The numbers greater than \(15\) are \(16,17,18,19,20\), so \(E_3^c\) has \(20-5=15\) outcomes. The numbers at most \(18\) are \(1\) through \(18\), so \(E_4^c\) has \(20-18=2\) outcomes: \(19\) and \(20\).

1. \(E_1^c\): The number is not prime; \(E_2^c\): the number is not a multiple of \(4\); \(E_3^c\): the number is at most \(15\); \(E_4^c\): the number is greater than \(18\). 2. \(E_1^c\): \(12\) outcomes; \(E_2^c\): \(15\) outcomes; \(E_3^c\): \(15\) outcomes; \(E_4^c\): \(2\) outcomes.

- Think about which outcomes remain after removing the outcomes in the stated event. - How do the words “all,” “none,” and “at least” change when an event is complemented? - What would have to happen to make the original statement false?

1. a) The original event requires every shot to hit the bullseye. Its complement occurs when at least one shot does not hit the bullseye. 2. b) The original event has zero bullseye hits. Its complement occurs when at least one shot hits the bullseye. 3. c) The original event requires both the first shot and the second shot to hit the bullseye. Its complement occurs when at least one of those first two shots does not hit the bullseye. 4. d) “At least five” means five or six bullseye hits. The remaining possibilities are zero through four bullseye hits, so the complement is at most four bullseye hits.

a) At least one shot does not hit the bullseye. b) At least one shot hits the bullseye. c) At least one of the first two shots does not hit the bullseye. d) At most four shots hit the bullseye.

- List the possible counts as \(0, 1, 2, \ldots, 20\). - Check the boundary carefully: Does “more than \(15\)” include \(15\)? - For a whole-number count, what is the only alternative to being even?

1. a) The possible numbers of defective batteries are the whole numbers from \(0\) through \(20\). Excluding exactly \(1\) leaves \(0\) or any number from \(2\) through \(20\). Thus, either no batteries or at least two batteries are defective. 2. b) “At most three” includes \(0, 1, 2, 3\). The remaining counts begin at \(4\), so at least four batteries are defective. 3. c) “More than \(15\)” includes \(16, 17, 18, 19, 20\). The remaining counts are \(0\) through \(15\), so at most \(15\) batteries are fully functional. 4. d) Every whole number is either even or odd. Therefore, the complement is that the number of defective batteries is odd.

a) Either no batteries or at least two batteries are defective. b) At least four batteries are defective. c) At most \(15\) batteries are fully functional. d) The number of defective batteries is odd.

- There are only two character types. If a character is not a digit, what must it be? - In part c, the complement occurs when even one of the two required conditions fails. - Express the number of letters or digits with inequalities to help identify the remaining cases.

1. a) The original event fails when at least one character is not a letter. Because every character is either a letter or a digit, the complement is that at least one character is a digit. 2. b) “At least one digit” means the number of digits is \(1\), \(2\), \(3\), or \(4\). The complement has zero digits, so all four characters are letters. 3. c) The original event requires both the first character and the last character to be digits. Its complement occurs when at least one of those two characters is a letter. 4. d) “At most two letters” includes zero, one, or two letters. The remaining possibilities are three or four letters, so the complement is at least three letters.

a) At least one character is a digit. b) All four characters are letters. c) At least one of the first and last characters is a letter. d) At least three characters are letters.

- What does it mean for an outcome to be outside an event? - How many outcomes are in the entire sample space, and how many are in the complement? - Recall the definition of a perfect square.

1. The perfect squares in the sample space are \(1,4,9\), so \(A=\{1,4,9\}\). 2. Removing those outcomes from \(S\) gives \(A^c=\{2,3,5,6,7,8,10\}\). 3. The numbers greater than \(7\) are \(8,9,10\), so \(B=\{8,9,10\}\). 4. Removing those outcomes from \(S\) gives \(B^c=\{1,2,3,4,5,6,7\}\). 5. The complement \(A^c\) has \(7\) favorable outcomes out of \(10\) equally likely outcomes. Therefore, \(P(A^c)=\frac{7}{10}=0.7\).

\(A=\{1,4,9\}\), \(A^c=\{2,3,5,6,7,8,10\}\) \(B=\{8,9,10\}\), \(B^c=\{1,2,3,4,5,6,7\}\) \(P(A^c)=\frac{7}{10}=0.7\)

- Which value gives the total number of members in set \(F\)? - How is the group that plays both soccer and tennis represented with set notation? - How can set \(F\) be separated into those who also play tennis and those who do not?

1. The number of soccer players is \(n(F)=75\), and the number who play both sports is \(n(F\cap T)=18\). 2. Members who play soccer but not tennis are in the set difference \(F\setminus T\). 3. Since the members in \(F\cap T\) are included in \(F\), subtract them: \(n(F\setminus T)=n(F)-n(F\cap T)\). 4. Substitute the values: \(n(F\setminus T)=75-18=57\).

\(n(F\setminus T)=n(F)-n(F\cap T)=75-18=57\), so \(57\) members play soccer but not tennis.

- In a Venn diagram, what does disjointness tell you about the overlap of the two circles? - Which elements are removed when you form a set difference? - What elements belong to \(N^c\) within the universal set \(U\)?

1. Disjoint sets have no common elements, so \(M\cap N=\emptyset\). 2. The set \(M\setminus N\) contains the elements of \(M\) that are not in \(N\). Since no element of \(M\) is in \(N\), \(M\setminus N=M\). 3. The complement \(N^c\) contains every element of \(U\) that is not in \(N\). Because every element of \(M\) lies outside \(N\), \(M\subseteq N^c\). Zofia's claim is correct.

a) \(M\cap N=\emptyset\) b) \(M\setminus N=M\), because removing elements of \(N\) removes nothing from \(M\). c) The claim is correct because every element of \(M\) is outside \(N\), so \(M\subseteq N^c\).

- First list every number that belongs to each set. - Which elements appear in both lists? - When forming a union, how many times should an element that appears in both sets be written? - How can factor pairs help you find all positive factors of \(24\)?

1. List the positive factors of \(24\): \(A=\{1,2,3,4,6,8,12,24\}\). 2. List the positive multiples of \(4\) less than \(30\): \(B=\{4,8,12,16,20,24,28\}\). 3. The elements common to both sets form the intersection: \(A\cap B=\{4,8,12,24\}\). 4. Combine all distinct elements from both sets to form the union: \(A\cup B=\{1,2,3,4,6,8,12,16,20,24,28\}\).

\(A\cap B=\{4,8,12,24\}\) \(A\cup B=\{1,2,3,4,6,8,12,16,20,24,28\}\)

- List the prime numbers between \(1\) and \(20\). - List the multiples of \(3\) between \(1\) and \(20\). - What must be true of an outcome in an intersection? - Should an outcome that belongs to both events be listed twice in the union?

1. The prime numbers from \(1\) through \(20\) are \(E_1=\{2,3,5,7,11,13,17,19\}\). 2. The multiples of \(3\) from \(1\) through \(20\) are \(E_2=\{3,6,9,12,15,18\}\). 3. The only outcome in both events is \(3\), so \(E_1\cap E_2=\{3\}\). 4. Combining all distinct outcomes gives \(E_1\cup E_2=\{2,3,5,6,7,9,11,12,13,15,17,18,19\}\).

\(E_1\cap E_2=\{3\}\) \(E_1\cup E_2=\{2,3,5,6,7,9,11,12,13,15,17,18,19\}\)

- For “and,” keep only values that satisfy both inequalities. - For “or,” include values that satisfy either inequality. - Pay attention to whether each endpoint is included before choosing a bracket or parenthesis.

1. For part a, \(x\) must satisfy both conditions. The overlap is \(3<x<5\), so the solution is \((3,5)\). 2. For part b, \(x\) may satisfy either condition. The two intervals overlap and together cover every value from \(-2\) through \(8\), including both endpoints, so the solution is \([-2,8]\).

a) \((3,5)\) b) \([-2,8]\)

- Use the visible rectangle as the full sample space when interpreting complements. - For an intersection, identify the region that belongs to both specified parts. - For a union, combine every region named by either set. - Describe each expression in words before deciding which part of its panel is included.

1. For \(A^c\cap B\), shade the part of \(B\) that lies outside \(A\), excluding the overlap. 2. For \(A\cup B^c\), shade all of \(A\) together with everything outside \(B\). The only unshaded region is the part of \(B\) outside \(A\). 3. For \((A\cup B)^c\), shade the part of \(S\) outside both circles.

a) The part of \(B\) that does not overlap \(A\). b) Every region except the part of \(B\) outside \(A\). c) The region inside \(S\) but outside both \(A\) and \(B\).

- For each description, identify which of the four regions in a two-set Venn diagram are included. - The word “or” usually corresponds to a union. - The word “and” usually corresponds to an intersection. - Recall what the superscript \(c\) means for an event.

1. “Both events occur” means the outcome is in the intersection, so the expression is \(P(A\cap B)\). 2. “At least one event occurs” means \(A\), \(B\), or both occur, so the expression is \(P(A\cup B)\). 3. “Neither event occurs” means both complements occur, so the expression is \(P(A^c\cap B^c)\). 4. “At most one event occurs” excludes only the case in which both occur. By De Morgan's law, \((A\cap B)^c=A^c\cup B^c\), so the expression is \(P(A^c\cup B^c)\).

1. \(P(A\cap B)\) 2. \(P(A\cup B)\) 3. \(P(A^c\cap B^c)\) 4. \(P(A^c\cup B^c)\)

- What must the probabilities of an event and its complement add to? - Express both probabilities in the same form before comparing them. - What value would make the two probabilities sum to \(1\)?

1. Convert the fraction to a decimal: \(P(B)=\frac{3}{8}=0.375\). 2. An event and its complement must have probabilities that sum to \(1\): \(P(B)+P(B^c)=1\). 3. The claimed values sum to \(0.375+0.65=1.025\). 4. Since \(1.025\ne1\), the claims cannot both be true. The correct complement probability would be \(1-0.375=0.625\).

No. The claimed probabilities sum to \(1.025\), not \(1\). If \(P(B)=\frac{3}{8}=0.375\), then \(P(B^c)=0.625\).

- For equally likely selections, compare the number of favorable outcomes with the total number of outcomes. - What happens when the caramel event does not occur? - How many chocolates remain after the caramel-filled chocolates are removed?

1. There are \(10\) caramel-filled chocolates out of \(25\), so \(P(C)=\frac{10}{25}=\frac{2}{5}=0.4\). 2. The complement \(C^c\) is the event that the selected chocolate does not have a caramel filling. 3. Use the complement rule: \(P(C^c)=1-P(C)=1-0.4=0.6\). 4. Equivalently, \(25-10=15\) chocolates have other fillings, and \(\frac{15}{25}=0.6\). 5. The caramel event and the combined event of all other fillings are complements, so their probabilities must add to \(1\).

a) \(P(C)=\frac{2}{5}=0.4=40\%\) b) \(C^c\) means the selected chocolate does not have a caramel filling, and \(P(C^c)=0.6=60\%\). c) All other fillings together form the complement of \(C\), so their combined probability is \(1-P(C)\).

- First list every ordered pair with a sum of \(8\). - Which of those pairs contain the number \(5\)? - What remains after removing the outcomes in \(B\) from event \(A\)? - Remember that the first and second rolls determine the order in each pair.

1. The ordered pairs whose sum is \(8\) are \(A=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}\). 2. The pairs in \(A\) that contain at least one \(5\) are \((3,5)\) and \((5,3)\). Therefore, \(A\cap B=\{(3,5),(5,3)\}\). 3. Remove the pairs containing a \(5\) from \(A\): \(A\setminus B=\{(2,6),(4,4),(6,2)\}\). 4. In words, \(A\setminus B\) is the event that the sum is \(8\) and neither roll is \(5\).

a) \(A=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}\) b) \(A\cap B=\{(3,5),(5,3)\}\) c) The sum is \(8\), and neither roll is \(5\). The outcomes are \(\{(2,6),(4,4),(6,2)\}\).

- Think of the total probability, \(1\), as being divided into equal ratio parts. - How many ratio parts belong to each outcome? - Which outcome is in both events? - What is the probability of an event that includes every possible outcome?

1. Add the ratio parts: \(2+1+4+2+3=12\). 2. Divide each part by \(12\): \(P(S_1)=\frac{2}{12}=\frac{1}{6}\), \(P(S_2)=\frac{1}{12}\), \(P(S_3)=\frac{4}{12}=\frac{1}{3}\), \(P(S_4)=\frac{2}{12}=\frac{1}{6}\), and \(P(S_5)=\frac{3}{12}=\frac{1}{4}\). 3. Add the probabilities of the outcomes in each event: \(P(A)=\frac{2+1+4}{12}=\frac{7}{12}\) and \(P(B)=\frac{4+2+3}{12}=\frac{9}{12}=\frac{3}{4}\). 4. Since \(A\cap B=\{S_3\}\), \(P(A\cap B)=\frac{1}{3}\). 5. Since \(A\cup B\) is the entire sample space, \(P(A\cup B)=1\).

a) \(P(S_1)=\frac{1}{6}\), \(P(S_2)=\frac{1}{12}\), \(P(S_3)=\frac{1}{3}\), \(P(S_4)=\frac{1}{6}\), \(P(S_5)=\frac{1}{4}\) b) \(P(A)=\frac{7}{12}\), \(P(B)=\frac{3}{4}\), \(P(A\cap B)=\frac{1}{3}\), \(P(A\cup B)=1\)

- Identify the values of \(X\) included in each original event. The complement contains all other possible values from \(0\) through \(25\). - Pay attention to the difference between “fewer than” and “at most,” and between “more than” and “at least.” - Think about which values remain after removing the given interval from all possible values of \(X\).

1. For \(E_1\), the values are \(X\le22\). The remaining possible values are greater than \(22\), so \(E_1^c\) is “More than \(22\) red balls are drawn,” or \(X>22\), equivalently \(X\ge23\). 2. For \(E_2\), the values are \(6\le X\le14\). The remaining values are below \(6\) or above \(14\), so \(E_2^c\) is “Fewer than \(6\) or more than \(14\) red balls are drawn,” or \(X<6\lor X>14\), equivalently \(X\le5\lor X\ge15\). 3. For \(E_3\), the values are \(X\ge1\). The only remaining possible value is \(X=0\), so \(E_3^c\) is “No red balls are drawn,” or \(X=0\).

a) \(E_1^c\): More than \(22\) red balls are drawn; \(X>22\), equivalently \(X\ge23\). b) \(E_2^c\): Fewer than \(6\) or more than \(14\) red balls are drawn; \(X<6\lor X>14\). c) \(E_3^c\): No red balls are drawn; \(X=0\).

- Consider every possible number of defective chips, from \(0\) through \(100\). - For an event that includes an interval of counts, identify the counts outside that interval. - To negate “no chips,” describe what must happen at least once.

1. Event \(E_1\) includes \(4\) through \(100\) defective chips. Its complement includes \(0\), \(1\), \(2\), or \(3\) defective chips: At most \(3\) chips are defective. 2. Event \(E_2\) includes \(10\) through \(20\) defective chips. Its complement includes counts below \(10\) or above \(20\): Fewer than \(10\) or more than \(20\) chips are defective. 3. Event \(E_3\) means there are \(0\) defective chips. Its complement includes every count from \(1\) through \(100\): At least one chip is defective.

a) \(E_1^c\): At most \(3\) chips are defective. b) \(E_2^c\): Fewer than \(10\) or more than \(20\) chips are defective. c) \(E_3^c\): At least one chip is defective.

- Compare the elements in the two sets and identify which letter they share. - For the union, include every letter from either set exactly once. - Check that every element you list belongs to the sample space \(S\).

1. The only letter that belongs to both \(A\) and \(B\) is \(A\). 2. Therefore, \(A\cap B=\{A\}\). 3. The letters that belong to at least one of the two sets are \(P\), \(L\), \(A\), and \(E\). 4. Therefore, \(A\cup B=\{P,L,A,E\}\).

\(A\cap B=\{A\}\) and \(A\cup B=\{P,L,A,E\}\).

- First write each event as a set of die outcomes. - Then check which outcomes belong to both events and which belong to at least one event.

1. The even outcomes are \(A=\{2,4,6\}\), and the outcomes greater than \(3\) are \(B=\{4,5,6\}\). 2. The outcomes common to both events are \(4\) and \(6\), so \(A\cap B=\{4,6\}\). 3. The outcomes in at least one event are \(2\), \(4\), \(5\), and \(6\), so \(A\cup B=\{2,4,5,6\}\).

\(A\cap B=\{4,6\}\), and \(A\cup B=\{2,4,5,6\}\).

- List both sets completely and in increasing order. - In the union, write each shared element only once.

1. The multiples of \(4\) are \(A=\{4,8,12,16,20\}\). 2. The numbers containing the digit \(1\) are \(B=\{1,10,11,12,13,14,15,16,17,18,19\}\). 3. The common elements are \(12\) and \(16\), so \(A\cap B=\{12,16\}\). 4. Combining all distinct elements gives \(A\cup B=\{1,4,8,10,11,12,13,14,15,16,17,18,19,20\}\).

\(A\cap B=\{12,16\}\), and \(A\cup B=\{1,4,8,10,11,12,13,14,15,16,17,18,19,20\}\).

- Which operation corresponds to elements in both sets, and which corresponds to elements in either set? - Does Mira's result omit elements that belong to only one of the sets?

1. The set \(\{3,5\}\) contains only the elements common to \(A\) and \(B\), so it is the intersection, not the union. 2. Therefore, \(A\cap B=\{3,5\}\). 3. The union contains every element that belongs to \(A\), \(B\), or both. 4. Thus, \(A\cup B=\{1,3,4,5,6,7\}\).

Mira confused the intersection with the union. The correct results are \(A\cap B=\{3,5\}\) and \(A\cup B=\{1,3,4,5,6,7\}\).

- First assign each film to event \(A\), event \(B\), both, or neither. - Distinguish “both features” from “at least one feature.”

1. The animated films are \(A=\{F_1,F_2,F_5\}\), and the films with no spoken dialogue are \(B=\{F_1,F_3,F_5\}\). 2. The films with both features are \(F_1\) and \(F_5\), so \(A\cap B=\{F_1,F_5\}\). 3. The films with at least one feature are \(F_1\), \(F_2\), \(F_3\), and \(F_5\), so \(A\cup B=\{F_1,F_2,F_3,F_5\}\).

\(A\cap B=\{F_1,F_5\}\), and \(A\cup B=\{F_1,F_2,F_3,F_5\}\).

- First identify every prime number in the sample space. - Then determine which outcomes belong to both events and which belong to at least one event.

1. The prime outcomes are \(A=\{2,3,5,7,11\}\), and the outcomes greater than \(8\) are \(B=\{9,10,11,12\}\). 2. The only common outcome is \(11\), so \(A\cap B=\{11\}\). 3. The outcomes in at least one event are \(2\), \(3\), \(5\), \(7\), \(9\), \(10\), \(11\), and \(12\). 4. Therefore, \(A\cup B=\{2,3,5,7,9,10,11,12\}\).

\(A\cap B=\{11\}\), and \(A\cup B=\{2,3,5,7,9,10,11,12\}\).

- Translate each event description into a set of outcomes. - Keep the order of the two tosses in mind.

1. The outcomes with heads on the first toss are \(A=\{HH,HT\}\), and the outcomes with exactly one tail are \(B=\{HT,TH\}\). 2. The only outcome common to both events is \(HT\), so \(A\cap B=\{HT\}\). 3. The outcomes in at least one of the events are \(HH\), \(HT\), and \(TH\), so \(A\cup B=\{HH,HT,TH\}\).

\(A\cap B=\{HT\}\), and \(A\cup B=\{HH,HT,TH\}\).

- Determine the two event sets independently. - Check each statement separately rather than selecting only one.

1. The perfect squares are \(A=\{4,9\}\), and the odd numbers are \(B=\{3,5,7,9\}\). 2. The only common element is \(9\), so statement I is correct. 3. The union is \(\{3,4,5,7,9\}\), so statement II is correct. 4. Statement III is false because \(4\) is not odd.

Statements I and II are correct; statement III is false.

- Compute the digit sum of each code systematically. - When forming the union, include each common element only once.

1. The codes with an even digit sum are \(A=\{11,13,15,17,19\}\). 2. The codes divisible by \(3\) are \(B=\{12,15,18\}\). 3. The only common code is \(15\), so \(A\cap B=\{15\}\). 4. Combining all distinct codes gives \(A\cup B=\{11,12,13,15,17,18,19\}\).

\(A\cap B=\{15\}\), and \(A\cup B=\{11,12,13,15,17,18,19\}\).

- Translate each symbol into ordinary language one operation at a time. - Remember that mathematical “or” includes the case in which both conditions are true.

1. The intersection \(A\cap B\) means the ticket is blue and has a star. 2. The union \(A\cup B\) means the ticket is blue, has a star, or has both features. 3. The event \(A\cap B^c\) means the ticket is blue and does not have a star.

a) The ticket is blue and has a star. b) The ticket is blue, has a star, or has both features. c) The ticket is blue and does not have a star.

- Check each listed outcome against the original condition. - An event and its complement cannot share an outcome.

1. The event is \(A=\{5,6\}\). 2. The complement cannot contain any outcome from \(A\). 3. Therefore, \(5\) does not belong in \(A^c\). 4. The correct complement is \(A^c=\{1,2,3,4\}\).

Leni incorrectly included \(5\). The correct set is \(A^c=\{1,2,3,4\}\).

- Pay close attention to whether the boundary value belongs to the original event. - Write the complete negation of the statement.

1. “No more than \(2\) minutes late” includes arriving on time and arriving up to and including \(2\) minutes late. 2. The complement begins strictly above \(2\) minutes late. 3. Therefore, statement II is correct.

Statement II: “The bus is more than \(2\) minutes late.”

- Count the total number of equal sectors shown in the spinner. - Which sectors lie outside event \(A\)? - For equally likely sectors, probability is favorable sectors divided by total sectors.

1. The diagram has \(10\) equal sectors. Three are green, so \(10-3=7\) sectors are not green and belong to \(A^c\). 2. Since the sectors are equal, \(P(A^c)=\frac{7}{10}=0.7\).

There are \(7\) sectors in \(A^c\), and \(P(A^c)=\frac{7}{10}=70\%\).

- Decide whether the boundary value belongs to the original event. - Then translate the negated inequality into individual outcomes.

1. Event \(A\) is described by \(X\le2\). 2. The negation is \(X>2\). 3. Within the sample space, \(A^c=\{3,4,5,6\}\).

\(A^c\): \(X>2\), and \(A^c=\{3,4,5,6\}\).

- Negate the entire statement “both sensors respond.” - Check that your wording covers every case outside event \(A\).

1. Event \(A\) occurs only when the first sensor and the second sensor both respond. 2. Negating the complete statement gives: at least one of the two sensors does not respond. 3. This wording includes the cases in which only the first fails, only the second fails, or both fail.

\(A^c\): “At least one of the two sensors does not respond.”

- Compare each proposal with the entire sample space. - Check both that the proposed complement has no overlap with \(A\) and that the two sets together cover \(S\).

1. The complement contains exactly the elements of \(S\) that are not in \(A\). 2. Those elements are \(b\), \(d\), and \(e\). 3. Therefore, only \(B_1=\{b,d,e\}\) equals \(A^c\). 4. Set \(B_2\) incorrectly includes \(c\), and set \(B_3\) incorrectly includes \(a\) and \(f\).

Only \(B_1=\{b,d,e\}\) equals \(A^c\).

- Remember that the sample space begins at \(0\). - First identify the outcomes outside \(A\), then compare their number with the total number of outcomes.

1. The event is \(A=\{0,1,2,3,4,5,6\}\). 2. The remaining outcomes form the complement: \(A^c=\{7,8,9\}\). 3. There are \(3\) favorable outcomes among \(10\) equally likely outcomes, so \(P(A^c)=\frac{3}{10}=0.3\).

\(A^c=\{7,8,9\}\), and \(P(A^c)=\frac{3}{10}=30\%\).

- First find the number of components outside event \(A\). - Then divide that number by the total sample size.

1. The number of components that do not pass is \(250-187=63\). 2. The sample proportion is \(\frac{63}{250}=0.252\). 3. As a percentage, this is \(25.2\%\).

There are \(63\) components in \(A^c\), representing \(0.252=25.2\%\) of the sample.

- Find the number of tickets outside event \(A\). - Convert that fraction to both requested forms.

1. The number of nonwinning tickets is \(200-46=154\). 2. Therefore, \(P(A^c)=\frac{154}{200}=0.77\). 3. As a percentage, \(P(A^c)=77\%\).

\(P(A^c)=0.77=77\%\).

- Decide whether \(10\) belongs to the original event. - Restrict the complement to the space numbers that are actually possible.

1. The original event is \(A=\{10,11,12\}\). 2. The negation of “at least \(10\)” is “less than \(10\).” 3. Within the given sample space, \(A^c=\{6,7,8,9\}\).

\(A^c\): “The space number is less than \(10\),” and \(A^c=\{6,7,8,9\}\).

- Consider every way in which the statement “both requirements are met” can be false. - Check whether an answer choice includes the case in which both requirements fail.

1. Event \(A\) requires both conditions to be true. 2. Its complement includes every case in which at least one condition fails. 3. This also includes the case in which both conditions fail. 4. Therefore, statement II is correct.

Statement II is correct.

- Treat the two possibilities as a complete partition of the outcomes. - When converting a decimal to a percentage, multiply by \(100\%\).

1. “The device has a startup failure or error” is the complement of “the device starts successfully without an error.” 2. Use the complement rule: \(1-0.905=0.095\). 3. As a percentage, \(0.095=9.5\%\).

The probability of a startup failure or error is \(0.095=9.5\%\).

- Check the boundary value between the two statements. - Complementary events must cover every possible outcome.

1. The integer \(5\) belongs to neither \(A\) nor \(B\). 2. Therefore, the two events do not cover all integers and are not complements. 3. The complement of “greater than \(5\)” is “less than or equal to \(5\).” 4. The corrected event is \(B\): “the integer is less than or equal to \(5\).”

No. The corrected event is \(B\): “The integer is less than or equal to \(5\).”

- What total must the two probabilities have? - Compare each proposed value with that total.

1. The probabilities of an event and its complement must sum to \(100\%\). 2. Compute the complement: \(100\%-35\%=65\%\). 3. Therefore, Jule is correct. 4. The check is \(P(A)+P(A^c)=1\).

Jule is correct: \(P(A^c)=65\%\). Check: the two probabilities must sum to \(100\%\).

- First list every number that belongs to event \(A\). - Then take the remaining elements of the sample space.

1. The numbers divisible by \(2\) are \(2,4,6,8\), and the numbers divisible by \(3\) are \(3,6,9\). 2. Therefore, \(A=\{2,3,4,6,8,9\}\). 3. The outcomes not in \(A\) are \(1\), \(5\), and \(7\). 4. Thus, \(A^c=\{1,5,7\}\).

\(A^c=\{1,5,7\}\).

- Check each number in the sample space for the prime-number and digit-sum conditions. - How does the number of outcomes in an event relate to its probability when all outcomes are equally likely? - To find a complement, identify every outcome in the sample space that is outside the event.

1. The prime numbers in the sample space are \(11,13,17,19\), so \(E=\{11,13,17,19\}\). 2. Removing those outcomes from \(S\) gives \(E^c=\{12,14,15,16,18,20\}\). 3. The numbers whose digits have a sum divisible by \(3\) are \(12\), \(15\), and \(18\). Therefore, \(F=\{12,15,18\}\). 4. Removing those outcomes from \(S\) gives \(F^c=\{11,13,14,16,17,19,20\}\). 5. Event \(E\) has \(4\) outcomes, while event \(F\) has \(3\). Because all \(10\) outcomes are equally likely, \(E\) is more likely than \(F\).

1. \(E=\{11,13,17,19\}\), \(E^c=\{12,14,15,16,18,20\}\) 2. \(F=\{12,15,18\}\), \(F^c=\{11,13,14,16,17,19,20\}\) 3. Event \(E\) is more likely because it has \(4\) outcomes, compared with \(3\) outcomes in \(F\).

- Convert each percentage to a count. - “Exactly one” excludes components in the intersection. - Separate the desired set into two disjoint parts.

1. Convert the percentages to counts: \(n(F_1)=0.08\cdot500=40\), \(n(F_2)=0.12\cdot500=60\), and \(n(F_1\cap F_2)=0.02\cdot500=10\). 2. Exactly one defect is the set \((F_1\setminus F_2)\cup(F_2\setminus F_1)\). 3. The number with only \(F_1\) is \(40-10=30\). 4. The number with only \(F_2\) is \(60-10=50\). 5. These two sets are disjoint, so the total is \(30+50=80\).

\(80\) components have exactly one of the two defects.

- Can you choose \(B\) so that it is disjoint from both \(A\) and \(C\), while \(A\) and \(C\) still overlap? - What does each given disjointness statement tell you about whether elements of \(A\) or \(C\) can lie in \(B\)? - In part c, identify exactly which elements are removed from the union.

1. The statement is false because disjointness is not transitive. For example, let \(A=\{1\}\), \(B=\{2\}\), and \(C=\{1\}\). Then \(A\cap B=\emptyset\) and \(B\cap C=\emptyset\), but \(A\cap C=\{1\}\). 2. Since neither \(A\) nor \(C\) contains an element of \(B\), every element of \(A\cup C\) is in \(B^c\). Therefore, \((A\cup C)\cap B^c=A\cup C\). 3. Removing \(B\) from \(A\cup B\cup C\) leaves the elements from \(A\) and \(C\), because both are disjoint from \(B\). Thus, \((A\cup B\cup C)\setminus B=A\cup C\).

a) False. For example, \(A=\{1\}\), \(B=\{2\}\), and \(C=\{1\}\) satisfy the given conditions, but \(A\) and \(C\) are not disjoint. b) \((A\cup C)\cap B^c=A\cup C\) c) \((A\cup B\cup C)\setminus B=A\cup C\)

- Consider a separate Venn diagram for each expression. - For the second expression, identify \(X^c\) and \(Y^c\) first, then combine them. - Which region remains unshaded in each diagram?

1. The intersection \(X\cap Y\) is the overlapping region of the two circles. Its complement, \((X\cap Y)^c\), includes all of \(S\) except that overlap. 2. The set \(X^c\) contains everything outside \(X\), and \(Y^c\) contains everything outside \(Y\). Their union contains every point that is outside at least one of the two sets. 3. The only points excluded from \(X^c\cup Y^c\) are those that lie in both \(X\) and \(Y\). 4. Therefore, the expressions describe the same region: \((X\cap Y)^c=X^c\cup Y^c\), which is De Morgan's law.

Yes. Both expressions include all of \(S\) except the overlap \(X\cap Y\), so \((X\cap Y)^c=X^c\cup Y^c\).

- Picture a smaller circle \(B\) completely inside a larger circle \(A\). - What is the intersection when every element of one set is already in the other? - Can an element be in \(B\) and outside \(A\) at the same time? - Think of \(A\setminus B\) as the part of the larger set left after removing the smaller set.

1. Since \(B\subsetneq A\), every element of \(B\) is also in \(A\). 2. Therefore, \(A\cap B=B\). 3. Since \(B\) adds no elements outside \(A\), \(A\cup B=A\). 4. No element can be both in \(B\) and outside \(A\), so \(A^c\cap B=\emptyset\). 5. The set \(A\cap B^c\) contains the elements of \(A\) that are not in \(B\), so \(A\cap B^c=A\setminus B\).

a) \(B\) b) \(A\) c) \(\emptyset\) d) \(A\setminus B\), the part of \(A\) outside \(B\)

- What does the superscript \(c\) mean for an event? - If you know the probability that at least one event occurs, how can you find the probability that neither occurs? - Which regions of a two-set Venn diagram are included in “at least one,” and which region is left out?

1. The event \(A\cup B\) means that at least one of the events \(A\) or \(B\) occurs. 2. The complement of “at least one occurs” is “neither event occurs.” 3. By the complement rule, \(P((A\cup B)^c)=1-P(A\cup B)\). 4. By De Morgan's law, \((A\cup B)^c=A^c\cap B^c\). 5. Therefore, \(P(A^c\cap B^c)=1-P(A\cup B)\). 6. Both sides represent the probability that neither \(A\) nor \(B\) occurs.

By De Morgan''s law, \((A\cup B)^c=A^c\cap B^c\). The complement rule then gives \(P(A^c\cap B^c)=P((A\cup B)^c)=1-P(A\cup B)\). Both sides mean the probability that neither event occurs.

- First identify the one Venn-diagram region that must not be included in \(E\). - Does “at most one” allow zero events to occur? - How can an event be described using the complement of the case that is excluded? - Under what condition is the intersection of two events impossible?

1. “At most one” includes the cases in which neither event occurs, only \(A\) occurs, or only \(B\) occurs. The only excluded case is that both events occur. 2. Since the intersection \(A\cap B\) is excluded, \(E=(A\cap B)^c\). 3. The three included cases can also be written as \(E=(A^c\cap B^c)\cup(A\cap B^c)\cup(A^c\cap B)\). 4. For \(E=S\), the excluded event must be empty. Therefore, \(A\cap B=\emptyset\), meaning \(A\) and \(B\) are disjoint.

a) Neither event occurs, only \(A\) occurs, or only \(B\) occurs. b) \(E=(A\cap B)^c\), or \(E=(A^c\cap B^c)\cup(A\cap B^c)\cup(A^c\cap B)\). c) \(E=S\) when \(A\cap B=\emptyset\), because then the excluded case can never occur.

- Think about which pairwise intersections guarantee that at least two events occur. - Can two events occur without the third event occurring? - How can you exclude the case in which all three events occur?

1. At least two events occur when \(A\) and \(B\), \(A\) and \(C\), or \(B\) and \(C\) occur. Thus, \(E=(A\cap B)\cup(A\cap C)\cup(B\cap C)\). 2. Rafael's claim is false. For example, an outcome may belong to \(A\cap B\) but not to \(C\). Then \(E\) occurs, but \(F=A\cap B\cap C\) does not. Therefore, \(F\subseteq E\), but \(E\) is not a subset of \(F\). 3. To represent exactly two events, remove the case in which all three occur: \(G=E\setminus(A\cap B\cap C)\). 4. Substituting the expression for \(E\) gives \(G=((A\cap B)\cup(A\cap C)\cup(B\cap C))\setminus(A\cap B\cap C)\).

a) \(E=(A\cap B)\cup(A\cap C)\cup(B\cap C)\) b) The claim is false. Two events can occur while the third does not. c) \(G=((A\cap B)\cup(A\cap C)\cup(B\cap C))\setminus(A\cap B\cap C)\)

- Describe the complement of at least one defective bulb. - Find the probability that one bulb is not defective. - Use the complement rule after calculating the probability that all ten bulbs are not defective.

1. The probability that one bulb is not defective is \(0.98\). 2. The complement of “at least one defective bulb” is “all ten bulbs are not defective.” 3. Thus, \(P(\text{at least one defective})=1-(0.98)^{10}\approx 0.1829\).

About \(0.1829\), or \(18.29\%\)

- State the logical opposite of “at least one.” - Multiply the probabilities for two independent acceptable components. - Subtract the complement probability from \(1\).

1. The complement is \(E^c\): Neither component is defective, so both components are acceptable. 2. The probability that one component is acceptable is \(1-0.05=0.95\). 3. By independence, \(P(E^c)=(0.95)^2=0.9025\). 4. Therefore, \(P(E)=1-P(E^c)=1-0.9025=0.0975\).

\(E^c\): Both components are acceptable. \(P(E)=0.0975\), or \(9.75\%\)

- Use a table or organized list so that every ordered pair appears once. - For each pair, determine both its sum and its product. - Sort each outcome according to whether it belongs to \(E\), \(F\), both, or neither. - What condition is true for outcomes in \(F\) but not in \(E\)?

1. The sample space is \(S=\{(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2)\}\). 2. The outcomes with an odd sum are \(E=\{(1,2),(2,1),(3,2),(4,1)\}\). 3. The outcomes with a product greater than \(4\) are \(F=\{(3,2),(4,2)\}\). 4. The common outcome is \(E\cap F=\{(3,2)\}\). 5. Removing the common outcome gives \(E\setminus F=\{(1,2),(2,1),(4,1)\}\) and \(F\setminus E=\{(4,2)\}\). 6. The event \(F\setminus E\) means the product is greater than \(4\) and the sum is even.

a) \(S=\{(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2)\}\) b) \(E\cap F=\{(3,2)\}\) c) \(E\setminus F=\{(1,2),(2,1),(4,1)\}\), and \(F\setminus E=\{(4,2)\}\). The latter event means the product is greater than \(4\) and the sum is even.

- The probabilities of all outcomes must add to \(1\). - First find the probability left after subtracting the known probability. - Divide the remaining probability according to the given ratio. - For the set operations, identify the outcomes in both events and the outcomes in at least one event.

1. The remaining probability is \(1-0.2=0.8\). 2. The ratio has \(1+2+5=8\) total parts, so each part represents \(0.8\div 8=0.1\). 3. Therefore, \(P(e_2)=0.1\), \(P(e_3)=0.2\), and \(P(e_4)=0.5\). 4. Add the probabilities of the outcomes in each event: \(P(E)=0.2+0.2=0.4\) and \(P(F)=0.1+0.2=0.3\). 5. Since \(E\cap F=\{e_3\}\), \(P(E\cap F)=0.2\). 6. Since \(E\cup F=\{e_1,e_2,e_3\}\), \(P(E\cup F)=0.2+0.1+0.2=0.5\).

a) \(P(e_1)=0.2\), \(P(e_2)=0.1\), \(P(e_3)=0.2\), \(P(e_4)=0.5\) b) \(P(E)=0.4\), \(P(F)=0.3\), \(P(E\cap F)=0.2\), \(P(E\cup F)=0.5\)

- The probabilities of all outcomes must add to \(1\). - Which outcome is not in event \(A\)? Use a complement. - A set with three elements has \(2^3\) subsets. - An event is a subset of the sample space. Add the probabilities of the outcomes it contains.

1. Since \(A^c=\{\omega_3\}\), \(P(\omega_3)=1-P(A)=1-0.55=0.45\). 2. Since \(B^c=\{\omega_1\}\), \(P(\omega_1)=1-P(B)=1-0.75=0.25\). 3. The outcome probabilities add to \(1\), so \(P(\omega_2)=1-0.25-0.45=0.30\). 4. Add the appropriate outcome probabilities for each subset of \(S\): \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), and \(P(S)=1\).

Individual outcomes: \(P(\omega_1)=0.25\), \(P(\omega_2)=0.30\), \(P(\omega_3)=0.45\) Events: \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), \(P(S)=1\)

- First identify which numbers from \(1\) through \(12\) satisfy each condition. - For a fair die, divide the number of favorable outcomes by the number of possible outcomes. - “Or” corresponds to a union. - “And not” corresponds to an intersection with a complement.

1. \(A=\{2,3,5,7,11\}\), \(B=\{1,2,3,4,6,12\}\), and \(C=\{9,10,11,12\}\). 2. Since the die is fair, each outcome has probability \(\frac{1}{12}\). Therefore, \(P(A)=\frac{5}{12}\), \(P(B)=\frac{6}{12}=\frac{1}{2}\), and \(P(C)=\frac{4}{12}=\frac{1}{3}\). 3. \(A\cup C=\{2,3,5,7,9,10,11,12\}\), so \(P(A\cup C)=\frac{8}{12}=\frac{2}{3}\). Also, \(B\cap C^c=\{1,2,3,4,6\}\), so \(P(B\cap C^c)=\frac{5}{12}\).

1. \(A=\{2,3,5,7,11\}\), \(B=\{1,2,3,4,6,12\}\), \(C=\{9,10,11,12\}\) 2. \(P(A)=\frac{5}{12}\), \(P(B)=\frac{1}{2}\), \(P(C)=\frac{1}{3}\) 3. \(P(A\cup C)=\frac{2}{3}\), \(P(B\cap C^c)=\frac{5}{12}\)

- What condition makes a number divisible by two given numbers at the same time? - What does a complement do to the original condition? - How are the probabilities of an event and its complement related? - Count the multiples of the relevant least common multiple.

1. A number divisible by both \(4\) and \(6\) must be a multiple of \(\operatorname{lcm}(4,6)=12\). 2. There are \(10\) multiples of \(12\) from \(1\) through \(120\), so \(P(E)=\frac{10}{120}=\frac{1}{12}\approx 0.0833\). 3. The complement is the event that the selected number is not divisible by both \(4\) and \(6\), equivalently, that it is not divisible by \(12\). 4. By the complement rule, \(P(E^c)=1-\frac{1}{12}=\frac{11}{12}\approx 0.9167\).

\(E^c\): “The selected number is not divisible by \(12\).” \(P(E)=\frac{1}{12}\approx 0.0833\), and \(P(E^c)=\frac{11}{12}\approx 0.9167\).

- List the numbers that satisfy each condition. - Check whether any number can satisfy both conditions. - Is \(1\) prime? - What is the negation of “A or B”?

1. The primes through \(50\) are \(\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}\), for a total of \(15\). 2. The perfect squares through \(50\) are \(\{1,4,9,16,25,36,49\}\), for a total of \(7\). 3. No perfect square in this list is prime, so the two sets are disjoint. Therefore, \(|E|=15+7=22\), and \(P(E)=\frac{22}{50}=0.44\). 4. The complement is “the number is neither prime nor a perfect square.” Thus \(P(E^c)=1-0.44=0.56\).

\(E^c\): “The number is neither prime nor a perfect square.” \(P(E)=0.44\), and \(P(E^c)=0.56\).

- Use the percentage with neither defect to find the probability of at least one defect. - Separate the union into components with \(M_1\) and components with only \(M_2\). - Which part of the union remains after the entire \(M_1\) event is removed?

1. The probability that a component has at least one defect is \(P(M_1\cup M_2)=1-0.90=0.10\). 2. The event \(M_1\cup M_2\) is the disjoint union of “has \(M_1\)” and “has \(M_2\) but not \(M_1\).” Therefore, \(P(M_2\cap M_1^c)=P(M_1\cup M_2)-P(M_1)=0.10-0.08=0.02\).

\(P(M_2\cap M_1^c)=0.02\), or \(2\%\).

- Count the total ordered outcomes for two die rolls. - Organize the possible sums in a table or grid. - Translate “at most,” “at least,” and “less than” carefully. - Identify the prime numbers from \(2\) through \(12\). - Count the grid entries that satisfy each event.

1. There are \(6\cdot6=36\) equally likely ordered outcomes. 2. For part a), the event is \(4\le X<7\). Sums \(4\), \(5\), and \(6\) occur in \(3\), \(4\), and \(5\) ways, respectively, so \(P(4\le X<7)=\frac{12}{36}=\frac{1}{3}\). 3. For part b), the prime sums are \(2\), \(3\), \(5\), \(7\), and \(11\). They occur in \(1+2+4+6+2=15\) ways, so \(P(X\in\{2,3,5,7,11\})=\frac{15}{36}=\frac{5}{12}\). 4. For part c), the event is \(X\le3\lor X>10\). Sums \(2\), \(3\), \(11\), and \(12\) occur in \(1+2+2+1=6\) ways, so the probability is \(\frac{6}{36}=\frac{1}{6}\).

a) \(P(4\le X<7)=\frac{1}{3}\) b) \(P(X\in\{2,3,5,7,11\})=\frac{5}{12}\) c) \(P(X\le3\lor X>10)=\frac{1}{6}\)

- Interpret the interval \([5, 8]\) as a set of possible integer values of \(S\). - For part b), identify sums satisfying both conditions. - For part c), consider counting the excluded sums and subtracting from \(1\). - Recall the symmetric pattern of counts for sums of two fair dice.

1. There are \(36\) equally likely ordered outcomes. 2. For part a), the event is \(5\le S\le8\). Sums \(5\), \(6\), \(7\), and \(8\) occur in \(4+5+6+5=20\) ways, so \(P(5\le S\le8)=\frac{20}{36}=\frac{5}{9}\). 3. For part b), the odd sums greater than \(8\) are \(9\) and \(11\). They occur in \(4+2=6\) ways, so \(P(S>8\land S\text{ is odd})=\frac{6}{36}=\frac{1}{6}\). 4. For part c), use the complement of \(S\in\{7, 11\}\). These sums occur in \(6+2=8\) ways, so \(P(S\notin\{7, 11\})=1-\frac{8}{36}=\frac{7}{9}\).

a) \(P(5\le S\le8)=\frac{5}{9}\) b) \(P(S>8\land S\text{ is odd})=\frac{1}{6}\) c) \(P(S\notin\{7, 11\})=\frac{7}{9}\)

- Separate the elements that appear only in the union from those specified in the intersection. - Check your final set against both given equations.

1. The elements \(b\), \(d\), and \(f\) are in the union but not in \(A\), so they must be in \(B\). 2. The intersection shows that \(c\) and \(e\) must also be in \(B\). 3. The elements \(a\) and \(g\) cannot be in \(B\), or they would also appear in the intersection. 4. Therefore, \(B=\{b,c,d,e,f\}\).

\(B=\{b,c,d,e,f\}\).

- Interpret each set equation separately. - For the die example, look for two groups that include all six outcomes without sharing any outcome.

1. The empty intersection means no outcome belongs to both events. 2. The union \(S\) means every outcome belongs to at least one of the events. 3. Therefore, the two events divide the sample space completely with no overlap; each event is the complement of the other. 4. For a die roll, one example is \(A=\{1,3,5\}\) and \(B=\{2,4,6\}\), representing odd and even outcomes.

The events are disjoint and together cover the entire sample space, so they are complements. For example, on one die roll, \(A=\{1,3,5\}\) and \(B=\{2,4,6\}\).

- Create one set for each feature. - For “exactly one,” exclude the tracks that have both features.

1. Let \(A=\{T_1,T_4,T_6\}\) represent the instrumental tracks and \(B=\{T_2,T_4,T_5,T_6\}\) represent the tracks shorter than \(3\) minutes. 2. The tracks with both features are \(A\cap B=\{T_4,T_6\}\). 3. The tracks with at least one feature are \(A\cup B=\{T_1,T_2,T_4,T_5,T_6\}\). 4. Remove the intersection from the union to find the tracks with exactly one feature: \(\{T_1,T_2,T_5\}\).

a) \(T_4,T_6\) b) \(T_1,T_2,T_4,T_5,T_6\) c) \(T_1,T_2,T_5\)

- Add all four regions to find the total number of visitors. - The union contains the three regions inside the two sets, while the complement intersection is the outside region.

1. The total number of visitors is \(48+27+39+16=130\). 2. The number who attend at least one show is \(48+27+39=114\). 3. Therefore, \(P(M\cup S)=\frac{114}{130}=\frac{57}{65}\approx0.8769\), or approximately \(87.7\%\). 4. The probability of attending neither show is \(P(M^c\cap S^c)=\frac{16}{130}=\frac{8}{65}\approx0.1231\), or approximately \(12.3\%\).

\(P(M\cup S)=\frac{57}{65}\approx87.7\%\) \(P(M^c\cap S^c)=\frac{8}{65}\approx12.3\%\)

- Find the two nonoverlapping parts separately. - Add only the probabilities for exactly one challenge.

1. The probability of solving only the code challenge is \(0.64-0.37=0.27\). 2. The probability of solving only the puzzle challenge is \(0.58-0.37=0.21\). 3. These events are disjoint, so the probability of exactly one challenge is \(0.27+0.21=0.48\).

\(0.48=48\%\)

- In the diagram, which part of the universal set belongs to neither circle? - What remains after all of set \(A\) is removed from the union? - If two sets have no common elements, where does one set lie relative to the complement of the other?

1. The complement \((A\cup B)^c\) contains all elements of \(U\) that are in neither \(A\) nor \(B\). In the diagram, this is the region inside the rectangle but outside both circles. 2. Removing \(A\) from \(A\cup B\) leaves \(B\), because \(A\) and \(B\) have no shared elements. Thus, \((A\cup B)\setminus A=B\). 3. Since \(A\cap B=\emptyset\), every element of \(A\) lies outside \(B\). Therefore, \(A\subseteq B^c\). 4. The union of a set with a superset is the superset, so \(A\cup B^c=B^c\). The equality is always true under the given condition.

a) \((A\cup B)^c\) contains the elements in neither \(A\) nor \(B\); it is the region inside \(U\) but outside both circles. b) \((A\cup B)\setminus A=B\) c) Yes. Since \(A\cap B=\emptyset\), \(A\subseteq B^c\), so \(A\cup B^c=B^c\).

- Partition the smartphone event into “both” and “smartphone only.” - Use the four disjoint regions of a two-event sample space. - For part c, apply De Morgan's law to identify the complement.

1. Since \(S=(S\cap T)\cup(S\cap T^c)\), \(P(S\cap T)=0.84-0.52=0.32\). 2. The tablet-only probability is \(P(S^c\cap T)=0.42-0.32=0.10\). 3. The probability of not owning a smartphone is \(P(S^c)=1-0.84=0.16\). 4. Therefore, the probability of owning neither device is \(P(S^c\cap T^c)=0.16-0.10=0.06\). 5. By De Morgan's law, \(S^c\cup T^c=(S\cap T)^c\). This event means that the respondent does not own both devices, or equivalently, is missing at least one of them. 6. Its probability is \(1-0.32=0.68\).

a) \(P(S\cap T)=0.32\) b) \(P(S^c\cap T^c)=0.06\) c) The event means that the respondent does not own both devices. \(P(S^c\cup T^c)=0.68\).

- Distinguish a condition on one trial from a condition that specifies an entire ordered outcome. - For “at least one,” consider the complementary event.

1. In part a, only the second component is specified. The other five components may be either defective or acceptable. The incorrect calculation instead describes the event that exactly the second component is defective. The correct probability is \(0.05\). 2. In part b, “at least one” includes one through six defective components. The incorrect expression calculates only the probability of exactly one defective component. Use the complement: \(P(\text{at least one defective})=1-(0.95)^6\approx 0.2649\).

a) The calculation incorrectly requires all other components to be acceptable. The correct probability is \(0.05\), or \(5\%\). b) The calculation includes only exactly one defective component. The correct probability is \(1-(0.95)^6\approx 0.2649\), or about \(26.49\%\).

- First identify the elements that must be in both sets. - Distribute the other elements of the union so that none becomes an additional shared element.

1. The elements \(2\) and \(6\) must belong to both sets. 2. Each of the remaining union elements \(1\), \(4\), and \(7\) must belong to at least one set, but not to both. 3. One choice is \(A=\{1,2,4,6\}\) and \(B=\{2,6,7\}\). 4. A second choice is \(A=\{1,2,6,7\}\) and \(B=\{2,4,6\}\). 5. Both choices have intersection \(\{2,6\}\) and union \(\{1,2,4,6,7\}\).

For example, \((A,B)=(\{1,2,4,6\},\{2,6,7\})\) and \((A,B)=(\{1,2,6,7\},\{2,4,6\})\).