GSEB Class 12 Statistics Solutions Chapter 1 Probability Exercise 1.1

GSEB Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.1

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.1

Question 1. State the sample space for the following random experiments:
(1) A balanced coin is thrown three times.
Answer:When a balanced coin is tossed three times, the total number of possible simple outcomes is \( n = 2^3 = 8 \).

• Head (H) appears on all three tosses, which is the outcome HHH.
• Head (H) appears on the first two tosses, and Tail (T) on the third, resulting in HHT.
• Head (H) on the first toss, Tail (T) on the second, and Head (H) on the third, forming HTH.
• Tail (T) on the first toss, followed by Head (H) on the next two, which is THH.
• Head (H) on the first toss, and Tail (T) on the next two tosses, creating HTT.
• Tail (T) on the first toss, Head (H) on the second, and Tail (T) on the third, leading to THT.
• Tail (T) on the first two tosses, then Head (H) on the third, giving TTH.
• Tail (T) appears on all three tosses, which is TTT.

Hence, the collection of all possible outcomes for throwing a balanced coin three times is: U = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}In simple words: The sample space lists every possible result when you flip a coin three times, showing combinations of heads and tails.

🎯 Exam Tip: For coin toss problems, remember that each toss has 2 outcomes (H or T). For 'n' tosses, the total outcomes are \(2^n\). Clearly list all unique combinations to define the sample space correctly.

 

Question 1.
(2) A balanced die with six sides and a balanced coin are tossed together.
Answer:The sample space for the random experiment of tossing a balanced six-sided die and a balanced coin together is expressed as follows: U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5,T), (6, T)} In each pair, the first value represents the number seen on the die, and the second value represents the outcome (Head or Tail) from the coin.In simple words: When you roll a die and flip a coin at the same time, the sample space includes all combinations of the die's number (1 to 6) with either a Head or a Tail.

🎯 Exam Tip: When combining two independent events (like a die roll and a coin toss), the total number of outcomes is the product of the number of outcomes for each individual event. For a die (6 outcomes) and a coin (2 outcomes), total outcomes = \(6 \times 2 = 12\).

 

Question 1.
(3) Two persons are to be selected from five persons a, b, c, d, e.
Answer:The total number of ways to choose two people from a group of five persons is calculated using combinations: \( 5C2 = \frac{5 \times 4}{2 \times 1} = 10 \) The sample space for this random experiment of selecting two persons from five persons (a, b, c, d, e) is: U = {(a, b), (a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, d), (c, e), (d, e)}In simple words: This lists all the different pairs you can pick from a group of five people, without caring about the order in which you pick them.

🎯 Exam Tip: Remember to use combinations (nCr) when the order of selection does not matter. If the order mattered (e.g., first person is president, second is vice-president), you would use permutations (nPr).

 

Question 2. Write the sample space for the marks (in integers) scored by a student appearing for an examination of 100 marks and state the number of sample points in it.
Answer:A student can score zero marks or any whole number mark from 1 to 100 in an examination. Thus, the sample space for the marks obtained by a student in an exam with 100 total marks is: U = {0, 1, 2, 3, ........., 100} The total number of possible marks (sample points) in this sample space U is 101.In simple words: The sample space here is every possible whole number score from 0 up to 100 that a student can get on a test.

🎯 Exam Tip: When defining a sample space for integers, make sure to include all valid integer values. For marks, often 0 is a possible score, so don't forget to include it, which increases the count by one compared to just 1 to 100.

 

Question 3. State the sample space for randomly selecting one minister and one deputy minister from four persons.
Answer:Let the four persons be represented by a, b, c, and d. First, one minister can be chosen from four persons in \( 4C1 \) ways. Then, one deputy minister can be chosen from the remaining three persons in \( 3C1 \) ways. So, the total number of possible selections is \( 4C1 \times 3C1 \) \( = 4 \times 3 \) \( = 12 \) Hence, the sample space for randomly choosing one minister and one deputy minister from four people is: U = {(a, b), (a, c), (a, d), (b, a), (b, c), (b, d), (c, a), (c, b), (c, d), (d, a), (d, b), (d, c)}. In each pair, the first person listed is the minister, and the second person is the deputy minister.In simple words: This lists all the unique ways to pick two people from four, where one person gets the 'minister' role and the other gets the 'deputy minister' role, meaning the order of selection matters.

🎯 Exam Tip: When roles or positions are assigned, the order of selection is important. This means you should use permutations or a sequential selection method (like \( n_1 \times n_2 \)) rather than combinations.

 

Question 4. A balanced coin in thrown in a random experiment till the first head is obtained. The experiment is terminated with a trial of first head. Write the sample space of this experiment and state whether it is finite or infinite.
Answer:A balanced coin is tossed repeatedly until the first head (H) appears. The possible outcomes of this random experiment are:

• Head (H) is obtained on the first toss.
• Tail (T) on the first toss, and Head (H) on the second toss (TH).
• Tail (T) on the first two tosses, then Head (H) on the third toss (TTH).
• Tail (T) on the first three tosses, then Head (H) on the fourth toss (TTTH).

This sequence of outcomes (TTTTH, TTTTTH, ...) can continue indefinitely because a head might not appear for a very long time. Therefore, the number of outcomes for this random experiment is indefinite. Thus, the sample space for throwing a balanced coin until the first head appears is: U = {H, TH, TTH, TTTH, ...} This sample space is infinite.In simple words: The sample space shows all the ways you could get the first "Heads" when flipping a coin, stopping as soon as "Heads" appears. Since "Heads" might take many tries, there's no end to the possibilities, making it an infinite list.

🎯 Exam Tip: A sample space is infinite if the number of possible outcomes cannot be counted or if the sequence of outcomes can continue indefinitely, like in experiments that stop based on a certain condition being met at an unknown point in time.

 

Question 5. Write the sample space for the experiment of randomly selecting three numbers from the first five natural numbers.
Answer:The first five natural numbers are 1, 2, 3, 4, 5. The number of ways to choose three numbers from these five is calculated using combinations: \( 5C3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \) Therefore, the sample space for randomly selecting three numbers from the first five natural numbers is: U = {(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5)}.In simple words: This is a list of all possible groups of three numbers you can pick from the numbers 1, 2, 3, 4, 5, where the order of numbers in a group doesn't matter.

🎯 Exam Tip: When "selecting" numbers or items without specifying order or distinct roles, always use combinations (\(nCr\)). Ensure all unique combinations are listed correctly.

 

Question 6. The sample space of a random experiment of selecting a number is U = {1, 2, 3,..., 20}. Write the sets showing the following events:
(1) The selected number is odd number.
(2) The selected number is divisible by 3.
(3) The selected number is divisible by 2 or 3.
Answer:Given sample space: U = {1, 2, 3, ........, 20}
(1) A = Event that the selected number is odd.
Therefore, A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
(2) B = Event that the selected number is divisible by 3.
Therefore, B = {3, 6, 9, 12, 15, 18}
(3) C = Event that the selected number is divisible by 2 or 3.
Therefore, C = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}In simple words: From a list of numbers 1 to 20, we identify three specific groups: numbers that are odd, numbers that can be divided by 3, and numbers that can be divided by either 2 or 3.

🎯 Exam Tip: For "divisible by X or Y", list all numbers divisible by X, then all numbers divisible by Y, and combine them without duplicates (A U B). For "divisible by X and Y", find numbers divisible by LCM(X, Y).

 

Question 7. One family is selected from the families having two children. The sex (male or female) of the children from this family is noted. State the sample space of this experiment and write the sets showing the following events:
(1) Event A₁ = One child is a female
(2) Event A₂ = At least one child is a female
Answer:Let 'B' represent a male child (boy) and 'G' represent a female child (girl). The sample space for observing the sex of two children in a family is: U = {(B, B), (B, G), (G, B), (G, G)}
(1) Event A₁ = The event that exactly one child is female.
A₁ = {(B, G), (G, B)}
(2) Event A₂ = The event that at least one child is female.
A₂ = {(B, G), (G, B), (G, G)}In simple words: First, we list all possible combinations of boy/girl for two children. Then, we find groups where there's exactly one girl, and groups where there's one or more girls.

🎯 Exam Tip: Clearly define symbols (e.g., B for boy, G for girl) at the start. "At least one" means one or more, while "exactly one" means precisely one. Listing all outcomes systematically helps avoid errors.

 

Question 8. Two six faced balanced dice are thrown simultaneously. State the sample space of this random experiment and hence write the sets showing the following events:
(1) Event A₁ = The sum of numbers on the dice is 7.
(2) Event A₂ = The sum of numbers on the dice is less than 4.
(3) Event A₃ = The sum of numbers on the dice is divisible by 3.
(4) Event A₄ = The sum of numbers on the dice is more than 12.
Answer:When two balanced six-sided dice are rolled simultaneously, the sample space consists of all possible pairs of outcomes (die 1, die 2): U = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Alternatively, U can be defined as: U= {(i, J); i, j = 1, 2, 3, 4, 5, 6}
(1) Event A₁ = The event where the sum of the numbers on the dice is 7.
A₁ = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
(2) Event A₂ = The event where the sum of the numbers on the dice is less than 4 (meaning the sum is 2 or 3).
A₂ = {(1, 1), (1, 2), (2, 1)}
(3) Event A₃ = The event where the sum of the numbers on the dice is divisible by 3.
A₃ = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}
(4) Event A₄ = The event where the sum of the numbers on the dice is more than 12.
Since the maximum sum is \(6+6=12\), it is not possible for the sum to be more than 12.
A₄ = { } OR \( \Phi \)In simple words: First, we list all 36 possible results when rolling two dice. Then, we find combinations that add up to 7, combinations that add up to less than 4, combinations whose sum is a multiple of 3, and combinations whose sum is greater than 12 (which is impossible).

🎯 Exam Tip: Systematically list all 36 outcomes for two dice rolls. For events, identify the specific pairs that satisfy the condition. Remember, an impossible event has an empty set as its outcome.

 

Question 9. Two numbers are selected at random from the first five natural numbers. The sum of two selected numbers is at least 6 is denoted by event A and the sum of two selected numbers is even is denoted by event B. Write the sets showing the following events and answer the given questions:
(1) U
(2) A
(3) B
(4) A U B
(5) A ∩ B
(6) A'
(7) A-B
(8) A' ∩ B
(9) Can it be said that the events A and B are mutually exclusive? Give reason.
(10) State the number of sample points in the sample space of this random experiment.
Answer:The first five natural numbers are 1, 2, 3, 4, 5. The total number of ways to select two numbers from these five is: \( 5C2 = \frac{5 \times 4}{2 \times 1} = 10 \)
(1) U = The sample space of selecting two numbers from the first five natural numbers.
U = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}
(2) A = The event that the sum of the two selected numbers is at least 6 (meaning the sum is 6 or more).
A = {(1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}
(3) B = The event that the sum of the two selected numbers is an even number.
B = {(1, 3), (1, 5), (2, 4), (3, 5)}
(4) A U B = The event that the sum of the two selected numbers is at least 6 OR an even number.
A U B = {(1, 3), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}
(5) A ∩ B = The event that the sum of the two selected numbers is at least 6 AND an even number.
A ∩ B = {(1, 5), (2, 4), (3, 5)}
(6) A' = The complementary event of A, meaning the sum of the two selected numbers is at most 6 (less than or equal to 6).
A' = U - A
A' = {(1, 2), (1, 3), (1, 4), (2, 3)}
(7) A - B = The event that the sum of the two selected numbers is at least 6 but not an even number. This means event A occurs, but event B does not occur.
A - B = A - (A ∩ B) = A ∩ B'
A - B = {(2, 5), (3, 4), (4, 5)}
(8) A' ∩ B = The event that the sum of the two selected numbers is not at least 6 (i.e., at most 6) AND is an even number. This means event A does not occur, but event B occurs.
A' ∩ B = B - (A ∩ B) = B - A
A' ∩ B = {(1, 3)}
(9) Can it be said that the events A and B are mutually exclusive? Give reason.
Answer: No. A ∩ B = {(1, 5), (2, 4), (3, 5)}. Since A ∩ B is not an empty set (\( \Phi \)), events A and B are not mutually exclusive. They have common outcomes.
(10) State the number of sample points in the sample space of this random experiment.
Answer:Number of sample points = 10.In simple words: We first find all pairs of numbers from 1 to 5. Then, we list groups of pairs based on their sum being at least 6, or their sum being an even number, or combinations of these conditions. We also check if events A and B share any outcomes.

🎯 Exam Tip: Carefully list the sample space (U) first. For set operations (union, intersection, complement, difference), apply the definitions precisely. Mutually exclusive events have no common outcomes (\( A \cap B = \Phi \)).

 

Question 10. Three female employees and two male employees are working in an office. One employee is selected from the employees of this office for training. The event that the employee selected for the training is a female is denoted by A and the event that this employee is a male is denoted by B. Find the sets showing the following events and answer the given questions:
(1) U
(2) A
(3) B
(4) A U B
(5) A ∩ B
(6) A' ∩ B
(7) Can it be said that the events A and B are mutually exclusive? Give reason.
(8) Can it be said that the events A and B are exhaustive? Give reason.
Answer:Let the three female employees be a, b, c and the two male employees be x, y. The sample space for selecting one employee is: U = {a, b, c, x, y}
(1) U = {a, b, c, x, y}
(2) A = The event that the employee selected for training is a female.
A = {a, b, c}
(3) B = The event that the employee selected for training is a male.
B = {x, y}
(4) A U B = The event that the employee selected for training is a female OR a male.
A U B = {a, b, c, x, y} = U
(5) A ∩ B = The event that the employee selected for training is both female AND male.
A ∩ B = { } = \( \Phi \)
(6) A' ∩ B = The event that the employee is not female (i.e., male) AND is male.
A' ∩ B = B - (A ∩ B)
A' ∩ B = B - A = B = {x, y}
(7) Can it be said that the events A and B are mutually exclusive? Give reason.
Answer: Yes. A ∩ B = \( \Phi \). Since their intersection is an empty set, events A and B are mutually exclusive events.
(8) Can it be said that the events A and B are exhaustive? Give reason.
Answer: Yes. A U B = U. Since the union of events A and B covers the entire sample space, events A and B are exhaustive events.In simple words: We list all employees and then categorize them by gender. Then we check for combinations like being female or male, both female and male (impossible), not female but male, and if these categories cover all employees and don't overlap.

🎯 Exam Tip: Events are mutually exclusive if they cannot happen at the same time (\(A \cap B = \Phi\)). Events are exhaustive if, together, they cover all possible outcomes in the sample space (\(A \cup B = U\)).

 

Question 11. One card is randomly drawn from a pack of 52 cards. If drawing a spade card is denoted by event A and drawing a card from ace to ten (non-face card) is denoted by B then write the sets showing the following events:
(1) U
(2) A
(3) B
(4) A U B
(5) A ∩ B
(6) B'
Answer:A standard pack of 52 cards includes 13 spade cards, 13 club cards, 13 heart cards, and 13 diamond cards. Let's denote the suits: S = Spade, C = Club, H = Heart, D = Diamond. Let's denote the ranks: A = Ace, 2-10 = Number cards, J = Jack, Q = Queen, K = King. Non-face cards are Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10 for each suit. The possible non-face cards for each suit are: Spade: S2, S3, S4, ..., S10, SA Club: C2, C3, C4, ..., C10, CA Heart: H2, H3, H4, ..., H10, HA Diamond: D2, D3, D4, ..., D10, DA The total number of ways to draw one card randomly from a pack of 52 cards is \( 52C1 = 52 \).
(1) U: The sample space for drawing one card randomly from a pack of 52 cards is expressed as follows: U = {SA, S2, S3, S4, ........ S10, SJ, SQ, SK, CA, C2, C3, C4, ........ C10, CJ, CQ, CK, HA, H2, H3, H4, ........ H10, HJ, HQ, HK, DA, D2, D3, D4, ........ D10, DJ, DQ, DK}
(2) Event A = Drawing a spade card.
A = {SA, S2, S3, S4, ........ S10, SJ, SQ, SK}
(3) Event B = Drawing a card from ace to ten (non-face card).
B = {SA, S2, S3, ........ S10, CA, C2, C3, ........ C10, HA, H2, H3, ........ H10, DA, D2, D3, ........ D10}
(4) Event A U B = Drawing a spade card OR a card from ace to ten.
A U B = {SA, S2, S3, S4, ........ S10, SJ, SQ, SK, CA, C2, C3, ........ C10, HA, H2, H3, ........ H10, DA, D2, D3, ........ D10}
(5) Event A ∩ B = Drawing a spade card AND a card from ace to ten.
A ∩ B = {SA, S2, S3, ........ S10}
(6) Event B' = Drawing a face card (or not a card from ace to ten).
B' = {SJ, SQ, SK, CJ, CQ, CK, HJ, HQ, HK, DJ, DQ, DK}In simple words: We list all 52 cards. Then, we find all spade cards (Event A), all cards from Ace to 10 (Event B), cards that are either spades or Ace-to-10, cards that are both spades and Ace-to-10, and finally, cards that are not Ace-to-10 (face cards).

🎯 Exam Tip: Clearly define the sample space (U) by listing all possible outcomes. For card problems, understand the composition of a standard deck (suits, ranks, face cards, non-face cards) to accurately identify events.

 

Question 12. The events A₁ and A₂ of a random experiment are as follows. Find the sets showing the union event A₁ U A₂ and intersection event A₁ ∩ A₂.
A₁ = {x|0 < x < 5}
A₂ = {x|- 1 < x < 3, x is an integer}
Answer:First, let's list the elements for each set: A₁ = {x|0 < x < 5}. This means x is a real number greater than 0 and less than 5. A₂ = {x| -1 < x < 3, x is an integer}. This means x is an integer greater than -1 and less than 3. Therefore, A₂ = {0, 1, 2}.
**Union event A₁ U A₂:** A₁ U A₂ = {x|0 < x < 5} \( \cup \) {0, 1, 2}
Since A₁ includes all real numbers between 0 and 5 (not including 0 and 5), and A₂ includes specific integers 0, 1, 2, the union will include all elements from A₁ and A₂.
\( \implies \) A₁ U A₂ = {x|0 \( \le \) x < 5}
**Intersection event A₁ ∩ A₂:** A₁ ∩ A₂ = {x|0 < x < 5} \( \cap \) {0, 1, 2}
This means we find elements that are common to both sets. The integers 1 and 2 from A₂ are also within the range of A₁. The integer 0 is in A₂ but not in A₁.
\( \implies \) A₁ ∩ A₂ = {1, 2}In simple words: For two sets of numbers, we first write out the numbers in each set. The union means combining all numbers from both sets. The intersection means finding only the numbers that are in both sets.

🎯 Exam Tip: Pay close attention to the variable type (integer, natural number, real number) and the inequality signs (<, >, ≤, ≥) when defining sets. Real numbers represent intervals, while integers are discrete points.

 

Question 13. The events A₁ and A₂ of a random experiment are as follows. Find the sets showing the union event A₁ U A₂ and intersection event A₁ ∩ A₂.
A₁ = {x|2 \( \le \) x < 6, x \( \in \) N}
A₂ = {x|3 < x < 9, x \( \in \) N}
Answer:First, let's list the elements for each set, remembering that N denotes natural numbers (positive integers starting from 1): A₁ = {x|2 \( \le \) x < 6, x \( \in \) N}. Therefore, A₁ = {2, 3, 4, 5}. A₂ = {x|3 < x < 9, x \( \in \) N}. Therefore, A₂ = {4, 5, 6, 7, 8}.
**Union event A₁ U A₂:** A₁ U A₂ = {2, 3, 4, 5} \( \cup \) {4, 5, 6, 7, 8}
\( \implies \) A₁ U A₂ = {2, 3, 4, 5, 6, 7, 8}
\( \implies \) A₁ U A₂ = {x|2 \( \le \) x \( \le \) 8, x \( \in \) N}
**Intersection event A₁ ∩ A₂:** A₁ ∩ A₂ = {2, 3, 4, 5} \( \cap \) {4, 5, 6, 7, 8}
\( \implies \) A₁ ∩ A₂ = {4, 5}
\( \implies \) A₁ ∩ A₂ = {x|x = 4, 5}In simple words: We list the whole numbers that fit the rules for set A1 and set A2. Then, for the union, we combine all unique numbers from both lists. For the intersection, we find only the numbers that appear in both lists.

🎯 Exam Tip: When dealing with natural numbers (N), ensure you start from 1 and include only positive integers. Clearly define the elements of each set before performing union or intersection operations.

 

Question 14. The sample space U of a random experiment and its event A are defined as follows. Find the complementary event A' of A.
U = {x|x = 0, 1, 2, ......... 10}
A = {x|x = 2, 4, 6}
Answer:Given sample space: U = {x|x = 0, 1, 2, ..., 10} This means U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Given event A: A = {x|x = 2, 4, 6} This means A = {2, 4, 6}. The complementary event A' consists of all elements in U that are not in A. A' = U - A A' = {0, 1, 3, 5, 7, 8, 9, 10}In simple words: We have a main list of numbers (U) and a smaller list of numbers (A). The complementary event A' is simply all the numbers from the main list that are not found in the smaller list.

🎯 Exam Tip: The complement of an event A (denoted A') includes all outcomes in the sample space (U) that are not in A. Always ensure A' + A = U and \(A \cap A' = \Phi\).

 

Question 15. The sample space U of a random experiment and Its event A are defined as follows. Find the complementary event A' of A.
U = {x|0 < x < 1}
A = {x|\(\frac{1}{2}\) \( \le \) x < 1}
Answer:Given sample space: U = {x|0 < x < 1}. This represents all real numbers strictly between 0 and 1. Given event A: A = {x|\(\frac{1}{2}\) \( \le \) x < 1}. This represents all real numbers from \( \frac{1}{2} \) up to, but not including, 1. The complementary event A' includes all elements in U that are not in A. A' = U - A Since U covers (0, 1) and A covers [\( \frac{1}{2} \), 1), A' will cover the remaining part of U. A' = {x|0 < x < \( \frac{1}{2} \)} The complementary event A' can be well explained by the following diagram:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख संख्या रेखा पर अंतराल (0, 1) को दिखाता है, जो U है। इसमें, A' (नीला क्षेत्र) अंतराल (0, \( \frac{1}{2} \)) को दर्शाता है, जबकि A (लाल क्षेत्र) अंतराल [\( \frac{1}{2} \), 1) को दर्शाता है। यह स्पष्ट करता है कि A' में 0 और \( \frac{1}{2} \) के बीच की सभी संख्याएँ शामिल हैं, जिसमें 0 शामिल नहीं है और \( \frac{1}{2} \) भी शामिल नहीं है। \[ \begin{array}{c} \textbf{U = 0 < x < 1} \\ \begin{array}{cc} \text{A'} & \text{A} \\ \hline \circ\text{-------}[\text{-------}\bullet\text{-------})\text{-------}\circ \\ \text{0} & \frac{1}{2} & \text{1} \\ \text{A' = 0 < x < } \frac{1}{2} & \text{A = } \frac{1}{2} \le \text{x < 1} \end{array} \end{array} \]In simple words: The main set (U) includes all numbers between 0 and 1. Set A includes numbers from 1/2 up to, but not including, 1. The complementary set A' includes all numbers in U that are not in A, meaning numbers between 0 and 1/2.

🎯 Exam Tip: When dealing with continuous intervals, visualize them on a number line. The complement is simply the part of the universal set that the event does not cover. Pay attention to whether endpoints are included or excluded (using <, >, ≤, ≥).