WAEC 2025 Mathematics Objective Answers: Step-by-Step Solutions

WAEC 2025 Maths Solutions: Step-by-Step Explanations

1. The volume of a cone of height 18 cm is 8,316 cm³. Find the base radius of the cone. [Take \(\pi = \frac{22}{7}\)]

   Solution

   Correct Option: A

   Using the formula \(V = \frac{1}{3}\pi r^2 h\):
   \(8316 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 18\)
   \(8316 = \frac{22}{7} \times r^2 \times 6\)
   \(r^2 = \frac{8316 \times 7}{22 \times 6} = \frac{58212}{132} = 441\)
   \(r = \sqrt{441} = 21\) cm.

2. A square has diagonal length of 10 cm. Find the perimeter of the square.

   Solution

   Correct Option: C

   Let the side be \(s\). Using Pythagoras' theorem: \(s^2 + s^2 = 10^2 \implies 2s^2 = 100 \implies s^2 = 50\).
   \(s = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) cm.
   Perimeter \(P = 4s = 4 \times 5\sqrt{2} = 20\sqrt{2}\) cm.

3. The height of 4 orange seedlings are: 2 cm, 5 cm, 7 cm and 10 cm. Calculate the variance.

   Solution

   Correct Option: C

   Mean (\(\mu\)) = \(\frac{2+5+7+10}{4} = \frac{24}{4} = 6\).
   Variance (\(\sigma^2\)) = \(\frac{\sum(x-\mu)^2}{N}\)
   \(\sigma^2 = \frac{(2-6)^2 + (5-6)^2 + (7-6)^2 + (10-6)^2}{4} = \frac{(-4)^2 + (-1)^2 + 1^2 + 4^2}{4}\)
   \(\sigma^2 = \frac{16+1+1+16}{4} = \frac{34}{4} = 8.5\).

4. Two towns X and Y are located at points (–2, –5) and (3, 7) respectively. Calculate the distance between the two towns.

   Solution

   Correct Option: C

   Distance \(D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
   \(D = \sqrt{(3 - (-2))^2 + (7 - (-5))^2} = \sqrt{(5)^2 + (12)^2}\)
   \(D = \sqrt{25+144} = \sqrt{169} = 13\) units.

5. Given that \((x^{-n})^4 = \frac{1}{x^6}\), find the value of \(n\).

   Solution

   Correct Option: C

   Using laws of indices: \((x^{-n})^4 = x^{-4n}\) and \(\frac{1}{x^6} = x^{-6}\).
   So, \(x^{-4n} = x^{-6}\). Equating the exponents: \(-4n = -6 \implies n = \frac{-6}{-4} = \frac{3}{2}\).

6. Solve: \(\frac{1}{3}(k - 4) - \frac{1}{2}(k + 1) < \frac{1}{6}\)

   Solution

   Correct Option: D

   Multiply all terms by the LCM (6): \(6[\frac{1}{3}(k-4)] - 6[\frac{1}{2}(k+1)] < 6[\frac{1}{6}]\)
   \(2(k-4) - 3(k+1) < 1\)
   \(2k - 8 - 3k - 3 < 1\)
   \(-k - 11 < 1 \implies -k < 12\).
   Multiply by -1 and reverse the inequality sign: \(k > -12\).

7. In a class of 42 students, 21 offer History and 28 offer Government...

   Solution

   Correct Option: C

   Total students \(n(H \cup G) = 42\), \(n(H) = 21\), \(n(G) = 28\).
   Number doing both: \(n(H \cap G) = n(H) + n(G) - n(H \cup G) = 21 + 28 - 42 = 7\).
   Number doing Government only = \(n(G) - n(H \cap G) = 28 - 7 = 21\).
   Probability = \(\frac{\text{Government only}}{\text{Total students}} = \frac{21}{42} = \frac{1}{2}\).

8. Consider the following statements: p: The weather is warm. q: The sun is shining...

   Solution

   Correct Option: D

   "The sun is shining" is \(q\). "The weather is warm" is \(p\). The logical connective "if and only if" is \(\leftrightarrow\). The statement "The sun is shining if and only if the weather is warm" is written as \(q \leftrightarrow p\).

9. The bearing of F from G is \(064^\circ\). What is the bearing of G from F?

   Solution

   Correct Option: C

   To find the back bearing for an angle less than 180°, you add 180° to it.
   Bearing of G from F = \(64^\circ + 180^\circ = 244^\circ\).

10. Ajoke will be 31 years in \(x\) years time. If Ajoke is 6 years older than Chioma, find an expression for Chioma's present age.

    Solution

    Correct Option: A

    Ajoke's present age (A): \(A + x = 31 \implies A = 31 - x\).
    Chioma's present age (C) is 6 years less than Ajoke's: \(C = A - 6\).
    Substitute the expression for A: \(C = (31 - x) - 6 = 25 - x\).

11. Simplify: \(\sqrt{200} - \sqrt{72}\)

    Solution

    Correct Option: B

    Simplify each surd: \(\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}\).
    \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\).
    Subtract: \(10\sqrt{2} - 6\sqrt{2} = 4\sqrt{2}\).

12. Simplify: \(\frac{2\frac{1}{3}+\frac{5}{2}-1}{12-11\frac{3}{4}}\)

    Solution

    Correct Option is B.

    Numerator: \(2\frac{1}{3} + \frac{5}{2} - 1 = \frac{7}{3} + \frac{5}{2} - 1 = \frac{14+15-6}{6} = \frac{23}{6}\).
    Denominator: \(12 - 11\frac{3}{4} = 12 - \frac{47}{4} = \frac{48-47}{4} = \frac{1}{4}\).
    Result: \(\frac{23/6}{1/4} = \frac{23}{6} \times 4 = \frac{92}{6} = \frac{46}{3} = {15}\frac{1}{3}\).

13. Given that \(x + y = 1\) and \(x + 3y = 5\), find the value of \((x^2 + 4xy + 3y^2)\).

    Solution

    Correct Option: B

    The expression \(x^2 + 4xy + 3y^2\) can be factored into \((x+y)(x+3y)\).
    We are given the values of the factors directly: \((x+y)=1\) and \((x+3y)=5\).
    Substitute the values: \((1)(5) = 5\).

14. Write 0.0364891 in three significant figures and express the answer in standard form.

    Solution

    Correct Option: D

    Rounding 0.0364891 to 3 significant figures gives 0.0365 (the 4 is rounded up because the next digit, 8, is 5 or greater).
    In standard form, this is \(3.65 \times 10^{-2}\).

15. The radius and height of a circular base cylinder are 8 cm and 14 cm respectively. Find its curved surface area. [Take \(\pi = \frac{22}{7}\)]

    Solution

    Correct Option: B

    Curved Surface Area = \(2\pi rh = 2 \times \frac{22}{7} \times 8 \times 14 = 2 \times 22 \times 8 \times 2 = 704\) cm².

16. An interior angle of a regular polygon is \(150^\circ\). How many sides has the polygon?

    Solution

    Correct Option: C

    The exterior angle = \(180^\circ - 150^\circ = 30^\circ\).
    Number of sides \(n = \frac{360^\circ}{\text{Exterior Angle}} = \frac{360^\circ}{30^\circ} = 12\).

17. Solve: \(d^2 - 4d - 96 = 0\)

    Solution

    Correct Option: D

    We need two numbers that multiply to -96 and add to -4. These are -12 and +8.
    The factored form is \((d-12)(d+8) = 0\).
    The solutions are \(d = 12\) or \(d = -8\).

18. The interior angles of a triangle are in the ratio 2 : 5 : 8. Find the difference between the smallest and largest angles.

    Solution

    Correct Option: D

    Sum of ratio parts = \(2+5+8=15\). The sum of angles in a triangle is 180°.
    Value per ratio part = \(\frac{180}{15}=12^\circ\).
    Smallest angle = \(2 \times 12^\circ = 24^\circ\). Largest angle = \(8 \times 12^\circ = 96^\circ\).
    Difference = \(96^\circ - 24^\circ = 72^\circ\).

19. The interior angles of a hexagon are \(107^\circ, (2x)^\circ, 150^\circ, 95^\circ, (2x - 15)^\circ\) and \(123^\circ\). Find the value of \(x\).

    Solution

    Correct Option: B

    Sum of interior angles of a hexagon = \((6-2) \times 180^\circ = 720^\circ\).
    Sum the given angles: \(107 + 2x + 150 + 95 + 2x - 15 + 123 = 720\).
    Combine terms: \(4x + 460 = 720 \implies 4x = 260 \implies x = 65\).

20. Two ball bearings have volumes 1.6 cm³ and 5.4 cm³. Find the ratio of their surface areas.

    Solution

    Correct Option: A

    Ratio of volumes = \(1.6 : 5.4\). To simplify, multiply by 10: \(16 : 54\). Divide by 2: \(8 : 27\).
    This is the ratio of lengths cubed (\(L^3\)).
    Ratio of lengths (\(L\)) = \(\sqrt[3]{8} : \sqrt[3]{27} = 2 : 3\).
    Ratio of surface areas (\(L^2\)) = \(2^2 : 3^2 = 4 : 9\).

21. Given that \(21_x = 14_5\), find the value of \(x\).

    Solution

    Correct Option: A

    Convert both numbers to base 10.
    Left side: \(21_x = 2(x^1) + 1(x^0) = 2x + 1\).
    Right side: \(14_5 = 1(5^1) + 4(5^0) = 5 + 4 = 9\).
    Set them equal: \(2x + 1 = 9 \implies 2x=8 \implies x=4\).

22. Two consecutive odd integers are such that the sum of 5 times the smaller and twice the bigger integer is 222. Find the value of the smaller integer.

    Solution

    Correct Option: A

    Let the smaller odd integer be \(n\), so the next is \(n+2\).
    The equation is \(5n + 2(n+2) = 222\).
    \(5n + 2n + 4 = 222 \implies 7n = 218 \implies n \approx 31.14\).
    Note: There is a likely typo in the question's total (222 should be 221). Testing the options, \(n=31\) gives \(5(31) + 2(33) = 155+66=221\), which is the closest value. Thus, 31 is the intended answer.

23. Kwakye, Sabina and Owusu shared an amount of $12,000.00...

    Solution

    Correct Option: C

    Kwakye's share = \(20\% \text{ of } 12000 = 0.20 \times 12000 = \$2400\).
    Remaining amount = \(12000 - 2400 = \$9600\).
    This is shared in ratio 5:3. Total parts = 8.
    Sabina's share = \(\frac{5}{8} \times 9600 = 5 \times 1200 = \$6000\).

24. The third and ninth terms of an Arithmetic Progression (AP) are 9 and -27 respectively. Find the fifth term.

    Solution

    Correct Option: B

    We have \(T_3 = a+2d = 9\) and \(T_9 = a+8d=-27\).
    Subtracting the first equation from the second: \(6d=-36 \implies d=-6\).
    Substitute \(d=-6\) into the first equation: \(a+2(-6)=9 \implies a=21\).
    The fifth term is \(T_5 = a+4d = 21+4(-6) = 21-24 =-3\).

25. If \(2\log_{x}({\frac{1-\sqrt{x}}{\sqrt{x}}})=0\), find the value of \(x\).

    Solution

    Correct Option: C

    First, divide both sides by 2: \(\log_{x}({\frac{1-\sqrt{x}}{\sqrt{x}}})=0\).
    For a logarithm to be equal to zero, its argument must be equal to 1.
    \(\frac{1-\sqrt{x}}{\sqrt{x}} = 1\).
    \(1-\sqrt{x} = \sqrt{x} \implies 1 = 2\sqrt{x}\).
    \(\sqrt{x} = \frac{1}{2}\).
    Squaring both sides: \(x = (\frac{1}{2})^2 = \frac{1}{4}\).

26. A cylinder and a cone have the same volume. If the height of the cone is 24 cm, find the height of the cylinder.

    Solution

    Correct Option: B

    Let \(V_{cyl}\) be the volume of the cylinder and \(V_{cone}\) be the volume of the cone. They have the same base radius \(r\).
    \(V_{cyl} = \pi r^2 h_c\) and \(V_{cone} = \frac{1}{3}\pi r^2 h_k\).
    Given \(V_{cyl} = V_{cone}\): \(\pi r^2 h_c = \frac{1}{3}\pi r^2 h_k\).
    Cancel \(\pi r^2\) from both sides: \(h_c = \frac{1}{3}h_k\).
    Given \(h_k = 24\) cm: \(h_c = \frac{1}{3}(24) = 8\) cm.

27. A variable \(P\) varies inversely as the square of \(Q\). If \(P = 5\) when \(Q = 6\), find \(Q\) when \(P = 1.8\).

    Solution

    Correct Option: B

    The relation is \(P = \frac{k}{Q^2}\).
    Find the constant \(k\): \(5 = \frac{k}{6^2} \implies 5 = \frac{k}{36} \implies k = 180\).
    The formula is \(P = \frac{180}{Q^2}\).
    Now find Q when P=1.8: \(1.8 = \frac{180}{Q^2} \implies Q^2 = \frac{180}{1.8} = 100 \implies Q=10\).

28. Factorize \(3 - 2x - x^2\).

    Solution

    Correct Option: D

    Rearrange the terms: \(-x^2 - 2x + 3\).
    Factor out -1: \(-(x^2 + 2x - 3)\).
    Factor the expression inside the parenthesis: \(-(x+3)(x-1)\).
    Distribute the negative sign into the second bracket: \((x+3)(-x+1)\), which is \((x+3)(1-x)\).

29. Make \(r\) the subject of the relation: \(\frac{1}{p} = \frac{b}{t} + \frac{c}{r}\)

    Solution

    The correct answer is D.

    Isolate the term with r: \(\frac{c}{r} = \frac{1}{p} - \frac{b}{t}\).
    Create a common denominator on the right side: \(\frac{c}{r} = \frac{t - pb}{pt}\).
    Invert both sides: \(\frac{r}{c} = \frac{pt}{t-pb}\).
    Multiply by c: \(r = \frac{pct}{t-pb}\).

30. In \(\triangle PQR\), \(|QR| = 2\) cm, \(\angle PRQ = 60^\circ\) and \(\angle PQR = 90^\circ\). Find \(|PR|\).

    Solution

    Correct Option: B

    In the right-angled triangle PQR, we can use trigonometric ratios.
    We have the angle \(\angle PRQ = 60^\circ\), the adjacent side \(|QR|=2\), and we want to find the hypotenuse \(|PR|\).
    Use the cosine ratio: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\).
    \(\cos(60^\circ) = \frac{2}{|PR|}\).
    Since \(\cos(60^\circ) = \frac{1}{2}\): \(\frac{1}{2} = \frac{2}{|PR|} \implies |PR| = 2 \times 2 = 4\) cm.

31. A sector of a circle of radius 21 cm subtends an angle of \(120^\circ\) at the centre...

    Solution

    Correct Option: C

    Length of arc = \(\frac{\theta}{360^\circ} \times 2\pi r\).
    \(L = \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{3} \times 2 \times 22 \times 3 = 44\) cm.

32. The population of a town increases by 2% every year. After 2 years, its population is 83,232...

    Solution

    Correct Option: A

    Let the original population be P. The formula for compound increase is \(A = P(1+R)^n\).
    \(83232 = P(1+0.02)^2 = P(1.02)^2 = P(1.0404)\).
    \(P = \frac{83232}{1.0404} = 80000\).

33. A profit of 8% was made when an article was sold for $500.00. At what price should it be sold to make a profit of 16%?

    Solution

    Correct Option: A

    First, find the Cost Price (CP): \(SP = CP \times (1 + \text{profit%}) \implies 500 = CP \times 1.08 \implies CP = \frac{500}{1.08}\).
    Now, calculate the new Selling Price (SP) for a 16% profit: \(SP = CP \times 1.16\).
    \(SP = \frac{500}{1.08} \times 1.16 \approx 537.037\).
    Rounding to two decimal places gives $537.04.

34. The sets \(M = \{x : 2 \leq x \leq 6\}\) and \(N = \{x : 4 \leq x \leq 8\}\) are subsets of \(\mu = \{x : 1 \leq x \leq 10, \text{where } x \text{ is an integer}\}\). Find \(M' \cap N'\).

    Solution

    Correct Option: C

    First, list the elements of each set:
    \(\mu = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\).
    \(M = \{2, 3, 4, 5, 6\} \implies M' = \{1, 7, 8, 9, 10\}\).
    \(N = \{4, 5, 6, 7, 8\} \implies N' = \{1, 2, 3, 9, 10\}\).
    The intersection \(M' \cap N'\) contains elements common to both M' and N': \(\{1, 9, 10\}\).

35. Given that \(\sin(x - 46)^\circ = \cos 62^\circ\), find the value of \(x\).

    Solution

    Correct Option: C

    Using the co-function identity that if \(\sin A = \cos B\), then the angles A and B are complementary, meaning \(A+B=90^\circ\).
    \((x - 46) + 62 = 90\).
    \(x + 16 = 90 \implies x = 90 - 16 = 74\).

36. Given that \(\log_4 16 = \log_y 36\), find the value of \(y\).

    Solution

    Correct Option: C

    First, evaluate the left side: \(\log_4 16 = 2\) because \(4^2 = 16\).
    The equation becomes \(2 = \log_y 36\).
    Convert this to exponential form: \(y^2 = 36\).
    Since the base of a logarithm must be positive, \(y=6\).

37. In the diagram, \(\angle PMQ = 34^\circ\) and \(\angle NQM = 28^\circ\). Find \(\angle QTN\).

    Solution

    Correct Option: D

    . \(\angle PNT= \angle TMP\) (Angle formed by same segment)

    \(\angle PNT = 34^\circ\)

    \(\angle QTN=180^\circ-\angle PNT-\angle TMP\) (Sum angles in a triangle )

    \(\angle QTN=180^\circ-28^\circ - 34^\circ = 118^\circ\)

    .

38. Find \(\angle MPN\).

    Solution

    Correct Option: A

    \(\angle MTN= 180 - \angle QTN\) (Sum of angles on a straight line)

    \(\angle MTN = 180 - 118 = 62^\circ\)

    \(\angle MTN=\angle MPN = 62^\circ\)

39. For what value of \(x\) is \(\frac{3x + 2}{2x + 1}\) undefined?

    Solution

    Correct Option: B

    A rational expression is undefined when its denominator is equal to zero.
    Set the denominator to zero and solve for x: \(2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}\).

40. In the diagram, P, Q, R and S are points on the circle with centre O. \(\angle QRS = (5x)^\circ\) and \(\angle QOS = (6x + 24)^\circ\). Find \(\angle QPS\).

    Solution

    Correct Option: A

    There are two key circle theorems at play:
    1. The angle at the center is twice the angle at the circumference subtended by the same arc. So, \(\angle QOS = 2 \times \angle QPS\).
    2. Opposite angles of a cyclic quadrilateral (PQRS) sum to 180°. So, \(\angle QPS + \angle QRS = 180^\circ\).
    From theorem 1: \(\angle QPS = \frac{\angle QOS}{2} = \frac{6x+24}{2} = 3x+12\).
    Now use theorem 2: \((3x+12) + (5x) = 180\).
    \(8x + 12 = 180 \implies 8x = 168 \implies x = 21\).
    The question asks for \(\angle QPS\), not x.
    \(\angle QPS = 3x+12 = 3(21)+12 = 63+12 = 75^\circ\).

41. In the diagram, O, R, S and T are points on the circle... \(\angle STR = 37^\circ\) and \(\angle QRT = 59^\circ\). Find \(\angle SPR\).

    Solution

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42. John sold an article for ₦105,000.00 at a loss of 4%. Find the cost price of the article.

    Solution

    Correct Option: D

    Selling at a 4% loss means the selling price (SP) is \(100\% - 4\% = 96\%\) of the cost price (CP).
    \(SP = 0.96 \times CP\).
    \(105000 = 0.96 \times CP\).
    \(CP = \frac{105000}{0.96} = 109375\). The cost price was ₦109,375.00.

43. In the diagram, \(SR \parallel UW\), \(\angle VUT = y\), \(\angle SXT = 45^\circ\) and \(\angle VTU = 20^\circ\). Find the angle marked \(x\).

    Solution

    Correct Option: A

    The angle marked \(x\) is \(\angle VXS\). This angle and \(\angle SXT\) form a linear pair on the straight line VT. Therefore, they sum to 180°.
    \(x + \angle SXT = 180^\circ\).
    \(x + 45^\circ = 180^\circ\).
    \(x = 180^\circ - 45^\circ = 135^\circ\).

44. Find the angle marked \(y\).

    Solution

    Correct Option: B

    Let the intersection of the lines VU and VT be A (the top vertex of the central triangle). The angle \(\angle SXT = 45^\circ\) is an exterior angle to this central triangle at vertex A.
    The exterior angle of a triangle is equal to the sum of the two opposite interior angles. The opposite interior angles are \(\angle ATU = 20^\circ\) and \(\angle AUT = y\).
    Therefore, \(45^\circ = y + 20^\circ\).
    \(y = 45^\circ - 20^\circ = 25^\circ\).

45. What is the gradient of the line \(7x - 5y = 3\)?

    Solution

    Correct Option: A

    To find the gradient, rearrange the equation into the slope-intercept form \(y = mx + c\), where \(m\) is the gradient.
    \(7x - 5y = 3\).
    \(-5y = -7x + 3\).
    Divide every term by -5: \(y = \frac{-7}{-5}x + \frac{3}{-5} \implies y = \frac{7}{5}x - \frac{3}{5}\).
    The gradient \(m\) is the coefficient of x, which is \(\frac{7}{5}\).

46. The Venn diagram illustrates the information about the sets P = {Primates}, C = {Smart People} and M = {My friends}. Which of the following is a valid conclusion from the diagram?

    Solution

    Correct Option: B

    The diagram shows three sets. The circle for "My friends" (M) overlaps with the circle for "Smart People" (C). The area of overlap (\(M \cap C\)) represents the members who are in both sets. Since this area exists, it means at least some of my friends are smart. Option B is the only valid conclusion.

47. The table shows the ages of students in a class. Find, correct to the nearest whole number, the average age of the students.

    Solution

    Correct Option: C

    To find the average (mean) age, we calculate \(\frac{\sum fx}{\sum f}\), where f is frequency and x is age.
    \(\sum f = 3+14+13+8+3 = 41\).
    \(\sum fx = (13 \times 3) + (14 \times 14) + (15 \times 13) + (16 \times 8) + (17 \times 3) = 39 + 196 + 195 + 128 + 51 = 609\).
    Average age = \(\frac{609}{41} \approx 14.85\).
    Correct to the nearest whole number, the average age is 15.

48. What is the median age?

    Solution

    Correct Option: B

    The total number of students (N) is 41. The median position is the \(\frac{N+1}{2}\)th value, which is \(\frac{41+1}{2} = 21\text{st}\) position.
    We use cumulative frequency to find where the 21st student falls:
    - Age 13: positions 1 to 3.
    - Age 14: positions 4 to (3+14) = 17.
    - Age 15: positions 18 to (17+13) = 30.
    The 21st student's age is 15. Therefore, the median age is 15.

49. The angle of elevation of the top P of a ladder PQ from its base Q, 12 m away from a wall is \(36^\circ\). Find correct to the nearest metre, the length of the ladder.

    Solution

    Correct Option: C

    This scenario forms a right-angled triangle where:
    - The distance from the wall is the adjacent side = 12 m.
    - The length of the ladder is the hypotenuse (L).
    - The angle of elevation is \(36^\circ\).
    We use the cosine ratio: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\).
    \(\cos(36^\circ) = \frac{12}{L}\).
    \(L = \frac{12}{\cos(36^\circ)} \approx \frac{12}{0.8090} \approx 14.83\) m.
    Correct to the nearest metre, the length is 15 m.

50. A fair dice is thrown once. Find the probability of obtaining an odd number or a prime number.

    Solution

    Correct Option: C

    The sample space S = {1, 2, 3, 4, 5, 6}.
    The set of odd numbers is A = {1, 3, 5}.
    The set of prime numbers is B = {2, 3, 5}. (Note: 1 is not a prime number).
    The event "odd or prime" is the union of these two sets: \(A \cup B = \{1, 3, 5\} \cup \{2, 3, 5\} = \{1, 2, 3, 5\}\).
    There are 4 favorable outcomes in this set.
    The total number of possible outcomes is 6.
    The probability is \(\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{4}{6} = \frac{2}{3}\).