CBSE Class 8 Mathematics Practice Worksheet Set 09

SECTION - A (1 mark for each)

Question 1. Write in standard form 0.0000507
Answer: \(5.07 \times 10^{-5}\)
In simple words: Move the decimal point 5 places to the right so that only one non-zero digit remains on the left.

Exam Tip: Moving the decimal to the right results in a negative exponent for base 10.

Question 2. If \(62y3\) is a multiple of 3, where \(y\) is a single digit then what should be the minimum and maximum value of \(y\)?
Answer: Sum of digits = \(6 + 2 + y + 3 = 11 + y\)
For \(62y3\) to be a multiple of 3, \(11 + y\) must be divisible by 3.
Possible values for \(11 + y\) are 12, 15, 18, etc.
If \(11 + y = 12 \implies y = 1\)
If \(11 + y = 15 \implies y = 4\)
If \(11 + y = 18 \implies y = 7\)
Minimum value of \(y = 1\), Maximum value of \(y = 7\)
In simple words: The sum of the digits must be a multiple of 3, so we test single-digit numbers for \(y\) starting from 0 to 9.

Exam Tip: A number is divisible by 3 if the sum of its digits is divisible by 3.

Question 3. What is the difference between Bar graph & Histogram?
Answer: A bar graph is a pictorial representation of data using bars of uniform width where gaps are left between consecutive bars. A histogram is a graphical representation of grouped frequency data in the form of contiguous (adjacent) rectangular bars with no gaps between them.
In simple words: Bar graphs have spaces between bars and represent categorical data, while histograms have no spaces and represent continuous data intervals.

Exam Tip: Always remember that histograms represent class intervals, so bars touch each other.

Question 4. If edge of a cube is doubled then its surface area becomes ____ times and the volume becomes ____ times.
Answer: Surface area becomes 4 times and volume becomes 8 times.
In simple words: If edge \(x\) becomes \(2x\), surface area \(6x^2\) becomes \(6(2x)^2 = 24x^2\) (4 times) and volume \(x^3\) becomes \((2x)^3 = 8x^3\) (8 times).

Exam Tip: Scale factor for surface area is square of the linear scale factor, and for volume it is the cube.

Question 5. What is the probability to get an even prime number in a single throw of a die.
Answer: Sample space \(S = \{1, 2, 3, 4, 5, 6\}\), Total outcomes = 6
Even prime number = 2, Favorable outcome = 1
Probability = \(\frac{1}{6}\)
In simple words: There is only one even prime number, which is 2, out of 6 possible outcomes on a die.

Exam Tip: 2 is the only even prime number in mathematics.

Question 6. TSA of a prism = LSA + 2 X ______
Answer: Base Area
In simple words: Total Surface Area includes the lateral surface area plus the area of the top and bottom bases.

Exam Tip: Memorize prism formulas carefully: \(\text{TSA} = \text{LSA} + 2 \times \text{Base Area}\).

Question 7. The point where two co-ordinate axes (x axis and y axis) intersect is called____
Answer: Origin
In simple words: The origin has coordinates \((0,0)\) where the x-axis and y-axis cross.

Exam Tip: Coordinates of the origin are always written as \((0,0)\).

Question 8. Write all possible outcomes in a toss of two coins together.
Answer: HH, HT, TH, TT
In simple words: The combinations are both heads, head and tail, tail and head, or both tails.

Exam Tip: Number of outcomes for \(n\) coins is \(2^n\). For 2 coins, it is \(2^2 = 4\).

Question 9. When number of edges in the base of a pyramid become very large (infinite) then it becomes a ______.
Answer: Cone
In simple words: As the polygonal base of a pyramid gets more and more sides, it approaches a circular base, forming a cone.

Exam Tip: Visualizing limits of shapes helps connect pyramids and cones.

Question 10. Multiply: \((2x + y)(3x - 2y)\)
Answer: \(= 2x(3x - 2y) + y(3x - 2y)\)
\(= 6x^2 - 4xy + 3xy - 2y^2\)
\(= 6x^2 - xy - 2y^2\)
In simple words: Use distributive multiplication to multiply each term of the first bracket by the second bracket.

Exam Tip: Combine like terms (\(-4xy + 3xy = -xy\)) carefully.

SECTION - B (3 marks for each)

Question 1. Write Eulers’s Formula, then find no. of faces in a solid if no. of vertices is 8 and no. of edges is 12.
Answer: Euler's Formula is: \(F + V - E = 2\) (or \(F + V = E + 2\))
Given, Vertices (\(V\)) = 8, Edges (\(E\)) = 12
\(F + 8 - 12 = 2\)
\(F - 4 = 2\)
\(F = 2 + 4 = 6\)
Number of faces = 6
In simple words: Use the formula connecting faces, vertices, and edges to find the missing number of faces.

Exam Tip: A cube or cuboid has 6 faces, 8 vertices, and 12 edges, which satisfies this formula.

Question 2. The data on the mode of transport used by 720 students are given below. Represent in a pie chart.

Answer: Total students = \(120 + 180 + 240 + 80 + 100 = 720\)
Central angle for each category = \(\left(\frac{\text{Value}}{\text{Total Value}}\right) \times 360^\circ\)
Bus: \(\left(\frac{120}{720}\right) \times 360^\circ = 60^\circ\)
Cycle: \(\left(\frac{180}{720}\right) \times 360^\circ = 90^\circ\)
Train: \(\left(\frac{240}{720}\right) \times 360^\circ = 120^\circ\)
Car: \(\left(\frac{80}{720}\right) \times 360^\circ = 40^\circ\)
Scooter: \(\left(\frac{100}{720}\right) \times 360^\circ = 50^\circ\)
In simple words: Convert each student count into a proportional angle out of \(360^\circ\) to draw the pie chart sectors.

Exam Tip: Verify that the sum of all central angles equals \(360^\circ\) (\(60^\circ + 90^\circ + 120^\circ + 40^\circ + 50^\circ = 360^\circ\)).

Question 3. Evaluate using suitable identity : \(10.5 \times 9.5\)
Answer: \(10.5 \times 9.5 = (10 + 0.5)(10 - 0.5)\)
Using identity \((a + b)(a - b) = a^2 - b^2\):
\(= (10)^2 - (0.5)^2\)
\(= 100 - 0.25\)
\(= 99.75\)
In simple words: Express the numbers as a sum and difference of the same base number and apply the difference of squares identity.

Exam Tip: Using algebraic identities saves time compared to standard multiplication.

Question 4. Find length of the attitude (altitude) of a rhombus if lengths of its two diagonals are 12 cm and 16 cm respectively.
Answer: Area of rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
Area = \(\frac{1}{2} \times 12 \times 16 = 96 \text{ cm}^2\)
Side of rhombus (\(s\)) = \(\sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}\)
\(s = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}\)
Area of rhombus also equals \(\text{Base} \times \text{Altitude}\)
\(96 = 10 \times \text{Altitude}\)
Altitude = \(\frac{96}{10} = 9.6 \text{ cm}\)
In simple words: First find the area and side length using diagonals, then equate the area formula to base times height to find the altitude.

Exam Tip: Diagonals of a rhombus bisect each other at right angles, forming right-angled triangles to find the side length.

Question 5. Find \(m\) if \(4^m = 8\)
Answer: Express both sides with base 2:
\((2^2)^m = 2^3\)
\(2^{2m} = 2^3\)
Equating exponents:
\(2m = 3\)
\(m = \frac{3}{2}\)
In simple words: Convert both bases to the same prime number (2) and equate their powers.

Exam Tip: If \(a^x = a^y\), then \(x = y\) when \(a \neq 0, 1, -1\).

Question 6. If 10 men do a work in 20 days. In how many days 20 men will do the same work.
Answer: This is an inverse proportion problem (\(M_1 \times D_1 = M_2 \times D_2\)):
\(10 \times 20 = 20 \times D_2\)
\(200 = 20 \times D_2\)
\(D_2 = \frac{200}{20} = 10 \text{ days}\)
In simple words: Double the number of workers means it takes half the time.

Exam Tip: More men mean fewer days, confirming inverse variation.

Question 7. Find \(A, B, C\) if:

\[ \begin{array}{r} AB\\ \times\ 5\\ \hline CAB \end{array} \]

Answer: Consider the multiplication \(\overline{AB} \times 5 = \overline{CAB}\):
Let \(AB = 10A + B\).
Multiplying by 5 means the product ends in 0 or 5. Since \(CAB\) has \(B\) as its last digit, \(B \times 5\) must end in \(B\).
Possible values for \(B\) are 0 or 5.
If \(B = 0\): \(\overline{A0} \times 5 = 50A\). For this to be a 3-digit number starting with \(C\) and ending in 0 where \(B=0\), let's test values of \(A\):
If \(A = 2\), \(20 \times 5 = 100\) (\(C=1, A=2, B=0\)).
Checking: \(20 \times 5 = 100\) (Matches \(CAB\), where \(C=1, A=2, B=0\)).
Therefore, \(A = 2\), \(B = 0\), \(C = 1\).
In simple words: Test digits for \(A\) and \(B\) such that the units digit multiplies correctly to satisfy the puzzle structure.

Exam Tip: Look at the units digit first in cryptarithms to narrow down possibilities.

SECTION - C (4 marks for each)

Question 1. Plot the following three vertices of a rectangle ABCD. \(A(0,0)\), \(B(5,0)\), \(C(5,3)\) then determine the fourth vertex with the help of graph.
Answer: Given vertices: \(A(0,0)\), \(B(5,0)\), \(C(5,3)\).
In a rectangle, opposite sides are equal and parallel.
The x-coordinate of \(D\) must align with \(A\), so \(x = 0\).
The y-coordinate of \(D\) must align with \(C\), so \(y = 3\).
Fourth vertex \(D = (0,3)\).
In simple words: Plot the given points on a coordinate grid and complete the rectangular box to find the missing corner.

Exam Tip: Opposite sides of a rectangle have matching coordinates in orthogonal directions.

Question 2. Draw a line graph for the following:-

Answer: To draw the line graph, take the side of the square along the X-axis and the perimeter along the Y-axis. Plot the points \((10, 40)\), \((20, 80)\), \((25, 100)\), \((30, 120)\), and \((40, 160)\), and connect them with straight line segments.
In simple words: Plot the pairs of numbers as points on graph paper and join them with straight lines.

Exam Tip: Choose an appropriate scale on both axes so the graph covers most of the grid.

Question 20. Factorise :- \(m^8 - n^8\)
Answer: \(m^8 - n^8 = ((m^4)^2 - (n^4)^2)\)
Using identity \(a^2 - b^2 = (a - b)(a + b)\):
\(= (m^4 - n^4)(m^4 + n^4)\)
\(= ((m^2)^2 - (n^2)^2)(m^4 + n^4)\)
\(= (m^2 - n^2)(m^2 + n^2)(m^4 + n^4)\)
\(= (m - n)(m + n)(m^2 + n^2)(m^4 + n^4)\)
In simple words: Repeatedly apply the difference of squares formula until no more factorization is possible.

Exam Tip: Remember that sum of squares like \((m^2 + n^2)\) cannot be factored further using real numbers.

Question 21. The marks obtained by 40 students of class VIII in an examination are given as followed: 16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 12, 15, 10, 17, 20, 8, 22, 18. Divide the data into five groups namely 0-5, 5-10, 10-15, 15-20 and 20-25. Then by choosing a suitable scale draw a histogram.
Answer:
Frequency distribution table:

To draw the histogram, represent class intervals along the X-axis and frequencies along the Y-axis, then draw adjacent rectangular bars for each interval.
In simple words: Group the numbers into ranges, count how many fall in each range, and draw touching bars on a graph.

Exam Tip: Always verify that the sum of frequencies equals the total number of students (40).

Question 22. Find LSA and TSA of a cube if its Volume is \(125 \text{ cm}^3\).
Answer: Volume of cube = \(a^3 = 125\)
Side (\(a\)) = \(\sqrt[3]{125} = 5 \text{ cm}\)
Lateral Surface Area (LSA) = \(4a^2 = 4(5)^2 = 4 \times 25 = 100 \text{ cm}^2\)
Total Surface Area (TSA) = \(6a^2 = 6(5)^2 = 6 \times 25 = 150 \text{ cm}^2\)
In simple words: Find the side length from the cube root of the volume, then use standard formulas for LSA and TSA.

Exam Tip: LSA excludes the top and bottom faces (4 faces), while TSA includes all 6 faces.

Question 23. Evaluate by using suitable identities:
a) \((99)^2\)
b) \(9.3 \times 9.7\)
Answer:
a) \((99)^2 = (100 - 1)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (100)^2 - 2(100)(1) + (1)^2\)
\(= 10000 - 200 + 1 = 9801\)
b) \(9.3 \times 9.7 = (9.5 - 0.2)(9.5 + 0.2)\)
Using \((a - b)(a + b) = a^2 - b^2\):
\(= (9.5)^2 - (0.2)^2\)
\(= 90.25 - 0.04 = 90.21\)
In simple words: Convert numbers into convenient sums or differences from round numbers and apply algebraic expansion identities.

Exam Tip: Choosing the right identity simplifies arithmetic calculations significantly.

Question 24. Define Prism and Pyramid with examples.
Answer:
Prism: A polyhedron whose base and top are congruent polygons and whose other faces (lateral faces) are parallelograms. Example: Triangular prism, rectangular prism (cuboid).
Pyramid: A polyhedron whose base is a polygon and whose lateral faces are triangles with a common vertex. Example: Square pyramid, triangular pyramid (tetrahedron).
In simple words: A prism has two identical flat ends and straight sides, while a pyramid has a flat base and triangular sides meeting at a top point.

Exam Tip: Remember that prisms have parallel bases while pyramids taper to a single apex.

Question 25. Find Volume of a cube if its TSA is \(150 \text{ cm}^2\).
Answer: Total Surface Area (TSA) = \(6a^2 = 150\)
\(a^2 = \frac{150}{6} = 25\)
Side (\(a\)) = \(\sqrt{25} = 5 \text{ cm}\)
Volume = \(a^3 = 5^3 = 125 \text{ cm}^3\)
In simple words: Divide the total surface area by 6 to find the area of one face, take its square root for the side length, then cube it for the volume.

Exam Tip: Keep units consistent (cm for length, cm\(^2\) for area, cm\(^3\) for volume).

Question 26. One of the parallel sides of a trapezium is 10 cm and its height is 4 cm. Find the length of the other side if its area is \(34 \text{ cm}^2\).
Answer: Area of trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Let the other parallel side be \(x\).
\(34 = \frac{1}{2} \times (10 + x) \times 4\)
\(34 = 2 \times (10 + x)\)
\(10 + x = \frac{34}{2} = 17\)
\(x = 17 - 10 = 7 \text{ cm}\)
In simple words: Substitute the known values into the trapezium area formula and solve the linear equation for the unknown side.

Exam Tip: Always include units in your final answer.

Question 27. Find Volume of a cylinder if its CSA and TSA are \(20\pi \text{ cm}^2\) and \(28\pi \text{ cm}^2\) respectively.
Answer: Curved Surface Area (CSA) = \(2\pi rh = 20\pi\)
Total Surface Area (TSA) = \(2\pi rh + 2\pi r^2 = 28\pi\)
Substitute CSA into TSA equation:
\(20\pi + 2\pi r^2 = 28\pi\)
\(2\pi r^2 = 28\pi - 20\pi = 8\pi\)
\(r^2 = \frac{8\pi}{2\pi} = 4 \implies r = 2 \text{ cm}\)
Now find height \(h\):
\(2\pi(2)h = 20\pi\)
\(4\pi h = 20\pi \implies h = 5 \text{ cm}\)
Volume of cylinder = \(\pi r^2 h = \pi(2)^2(5) = 20\pi \text{ cm}^3\)
In simple words: Subtract CSA from TSA to find the area of the two circular bases, determine the radius and height, then calculate the volume.

Exam Tip: TSA is simply CSA plus the area of the top and bottom circular bases (\(2\pi r^2\)).

Question 28. A 5m 60cm high pole casts shadow of length 3m 20cm. Find at the same time the length of a shadow cast by another pole 10m 50cm high.
Answer: Convert all dimensions to cm:
Height of first pole = \(5\text{ m } 60\text{ cm} = 560 \text{ cm}\)
Shadow of first pole = \(3\text{ m } 20\text{ cm} = 320 \text{ cm}\)
Height of second pole = \(10\text{ m } 50\text{ cm} = 1050 \text{ cm}\)
Let shadow of second pole be \(x\).
Using direct proportion (\(\frac{h_1}{s_1} = \frac{h_2}{s_2}\)):
\(\frac{560}{320} = \frac{1050}{x}\)
\(\frac{7}{4} = \frac{1050}{x}\)
\(7x = 1050 \times 4 = 4200\)
\(x = \frac{4200}{7} = 600 \text{ cm} = 6 \text{ m}\)
In simple words: Set up a direct proportion between the heights and shadow lengths after converting everything to centimeters.

Exam Tip: Always convert mixed units into a single unit (like cm) before solving ratios.

Question 29. Simplify and express the result in power notation with positive exponent:
(i) \((-3)^4 \times \left(\frac{5}{3}\right)^4\)
(ii) \((3^{-7} \div 3^{-10}) \times 3^{-5}\)
Answer:
(i) \((-3)^4 \times \left(\frac{5}{3}\right)^4 = 3^4 \times \frac{5^4}{3^4} = 5^4\)
(ii) \((3^{-7} \div 3^{-10}) \times 3^{-5} = 3^{-7 - (-10)} \times 3^{-5} = 3^3 \times 3^{-5} = 3^{3 + (-5)} = 3^{-2} = \frac{1}{3^2}\)
In simple words: Use exponent laws to combine terms with the same base and convert any negative exponents into positive ones by taking reciprocals.

Exam Tip: Remember that \((-a)^n = a^n\) when \(n\) is an even integer.

SECTION - D (7 marks for each)

Question 30. (a) Factorise then divide: \(\frac{-156y^3(36y^2 - 64)}{104y^2(6y + 8)}\)
(b) Factorise: \(-16a^2 - 25b^2 + 60bc - 36c^2\)
Answer:
(a) Numerator: \(-156y^3(36y^2 - 64) = -156y^3 \cdot 4(9y^2 - 16) = -624y^3(3y - 4)(3y + 4)\)
Denominator: \(104y^2(6y + 8) = 104y^2 \cdot 2(3y + 4) = 208y^2(3y + 4)\)
Divide:
\(\frac{-624y^3(3y - 4)(3y + 4)}{208y^2(3y + 4)} = -3y(3y - 4) = -9y^2 + 12y\)
(b) \(-16a^2 - 25b^2 + 60bc - 36c^2 = -16a^2 - (25b^2 - 60bc + 36c^2)\)
\(= -(4a)^2 - ((5b)^2 - 2(5b)(6c) + (6c)^2)\)
\(= -(4a)^2 - (5b - 6c)^2\)
Using difference of squares \(X^2 - Y^2 = (X - Y)(X + Y)\):
\(= [4a - (5b - 6c)][4a + (5b - 6c)]\)
\(= (4a - 5b + 6c)(4a + 5b - 6c)\)
In simple words: Factor terms completely using algebraic identities and common factors before performing division or regrouping.

Exam Tip: Group negative signs carefully when completing squares for polynomial factorisation.

Question 31. If two dice are thrown together, find all possible outcomes (i.e. sample space). Further find the probability to get the sum of digits on their upper faces as to be (a) exactly 8 (b) at least 8 (c) at most 8.
Answer: Total outcomes = \(6 \times 6 = 36\).
(a) Sum exactly 8: \((2,6), (3,5), (4,4), (5,3), (6,2)\) -> 5 outcomes.
Probability = \(\frac{5}{36}\)
(b) Sum at least 8 (sum \(\geq\) 8):
Sum 8: 5 outcomes
Sum 9: \((3,6), (4,5), (5,4), (6,3)\) -> 4 outcomes
Sum 10: \((4,6), (5,5), (6,4)\) -> 3 outcomes
Sum 11: \((5,6), (6,5)\) -> 2 outcomes
Sum 12: \((6,6)\) -> 1 outcome
Total favorable = \(5 + 4 + 3 + 2 + 1 = 15\) outcomes.
Probability = \(\frac{15}{36} = \frac{5}{12}\)
(c) Sum at most 8 (sum \(\le\) 8):
Total outcomes = 36. Outcomes with sum \(> 8\) are sums 9, 10, 11, 12 which total \(4 + 3 + 2 + 1 = 10\) outcomes.
Favorable = \(36 - 10 = 26\) outcomes.
Probability = \(\frac{26}{36} = \frac{13}{18}\)
In simple words: List or count pairs that satisfy the sum condition and divide by 36 total outcomes.

Exam Tip: For "at most" or "at least", finding the complementary event can save time.

Question 32. A rectangular sheet of length 176 cm and width 88 cm is rolled along its length and thus formed a cylinder. Find its volume. Now it is rolled along its width and turned into a cylinder again. Compare the volume of cylinder so formed in both case and find in which case the volume is more and by how much.
Answer:
Case 1: Rolled along length (Length = circumference, Width = height)
Circumference \(2\pi r_1 = 176 \implies 2 \times \frac{22}{7} \times r_1 = 176 \implies r_1 = \frac{176 \times 7}{44} = 28 \text{ cm}\)
Height \(h_1 = 88 \text{ cm}\)
Volume \(V_1 = \pi r_1^2 h_1 = \frac{22}{7} \times (28)^2 \times 88 = \frac{22}{7} \times 784 \times 88 = 216480 \text{ cm}^3\)
Case 2: Rolled along width (Width = circumference, Length = height)
Circumference \(2\pi r_2 = 88 \implies 2 \times \frac{22}{7} \times r_2 = 88 \implies r_2 = \frac{88 \times 7}{44} = 14 \text{ cm}\)
Height \(h_2 = 176 \text{ cm}\)
Volume \(V_2 = \pi r_2^2 h_2 = \frac{22}{7} \times (14)^2 \times 176 = \frac{22}{7} \times 196 \times 176 = 108240 \text{ cm}^3\)
Comparison: \(V_1 > V_2\). Volume is more when rolled along its length.
Difference = \(216480 - 108240 = 108240 \text{ cm}^3\)
In simple words: The dimension along which the sheet is rolled becomes the base circumference, and the other dimension becomes the height.

Exam Tip: Volume depends more heavily on the radius than the height because \(r\) is squared in \(\pi r^2 h\).

Question 33. If 8 men can complete a piece of work in 6 days. Find in how many days 12 men can complete twice of that work.
Answer: Using worker-day formula (\(\frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2}\)):
\(\frac{8 \times 6}{1} = \frac{12 \times D_2}{2}\)
\(48 = 6 \times D_2\)
\(D_2 = \frac{48}{6} = 8 \text{ days}\)
In simple words: Adjust for both the number of workers and the amount of work using proportional formulas.

Question 34. Diagram of the below picture frame has outer dimensions = \(24 \text{ cm} \times 28 \text{ cm}\) and inner dimensions = \(16 \text{ cm} \times 20 \text{ cm}\). Find the area of each section of the frame, if the width of each section is same.
Answer: Outer dimensions = \(24 \times 28\), Inner dimensions = \(16 \times 20\)
Width of frame border = \(\frac{28 - 20}{2} = 4 \text{ cm}\)
The frame consists of 4 sections: 2 trapeziums on top and bottom, and 2 trapeziums on left and right.
Area of top/bottom trapezoidal sections = \(2 \times \left[ \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \right]\)
Top section parallel sides are 24 and 16, height is 4:
Area of one such section = \(\frac{1}{2} \times (24 + 16) \times 4 = \frac{1}{2} \times 40 \times 4 = 80 \text{ cm}^2\)
Side sections parallel sides are 28 and 20, height is 4:
Area of side section = \(\frac{1}{2} \times (28 + 20) \times 4 = \frac{1}{2} \times 48 \times 4 = 96 \text{ cm}^2\)
In simple words: Treat each border piece as a trapezium with known parallel sides and width as height.

Exam Tip: Symmetric sections opposite to each other have identical areas.

Question 35. If three coins are thrown together then write total possible outcomes and the probability to get (a) exactly three heads (b) at least three heads (c) at most two heads.
Answer: Total outcomes = \(2^3 = 8\)
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(a) Exactly three heads (HHH): 1 outcome -> Probability = \(\frac{1}{8}\)
(b) At least three heads (3 or more heads): HHH -> 1 outcome -> Probability = \(\frac{1}{8}\)
(c) At most two heads (0, 1, or 2 heads): All outcomes except HHH -> 7 outcomes -> Probability = \(\frac{7}{8}\)
In simple words: Write out all 8 combinations of 3 coin tosses and count the ones matching each condition.

Exam Tip: "At most two heads" is the complement of "exactly three heads", so its probability is \(1 - \frac{1}{8} = \frac{7}{8}\).