USTET Mathematics — Statistics & ProbabilityExam Answer Templates
How to answer Statistics & Probability questions on the USTET — a set of templates you can apply to any question University of Santo Tomas throws at you in the Mathematics subtest. Built from analysis of recent USTET 2026 papers.
Exam context
University of Santo Tomas runs the University of Santo Tomas Entrance Test on Early Q4 2026. Its Mathematics section sits under a "Core section" weighting, and Statistics & Probability is the 8th chapter in the 9-chapter USTET Mathematics rotation. The USTET passing mark is Competitive overall score, and the most recent 2026 paper drew about a meaningful share of questions from Mathematics.
Statistics & Probability - Exam answer templates
Proper answer writing is crucial for maximizing your score in Statistics & Probability questions. The key to success lies in showing your complete working, using correct mathematical notation, and presenting solutions in a logical, step-by-step format. Examiners award marks for method as well as accuracy, so even if your final answer is incorrect, you can still earn partial marks by showing the correct approach. This chapter covers various question types from basic calculations to complex probability problems, requiring different answer structures for different mark allocations.
Templates
Find the mean of the following data: 12, 15, 18, 22, 25, 30
Marks
2
Topic
Measures of Central Tendency
Difficulty
easy
Template Id
T1
Examiner Tip
Always show the sum calculation explicitly - don't just write the final division
Model Answer
Given: Data set = {12, 15, 18, 22, 25, 30} To Find: Mean Solution: Mean = Sum of all observations / Number of observations Mean = (12 + 15 + 18 + 22 + 25 + 30) / 6 Mean = 122 / 6 Mean = 20.33 Therefore, the mean is 20.33
Question Type
short_answer
Answer Structure
- Line 1: Write Given and To Find [0.5 marks]
- Line 2: State the formula for mean [0.5 marks]
- Line 3: Substitute values and calculate sum [0.5 marks]
- Line 4: Final division and answer [0.5 marks]
Scoring Breakdown
Marks
1
Criteria
Correct formula identification and setup
Marks
1
Criteria
Accurate calculation and final answer
Common Mark Deductions
- Not showing the addition step
- Calculation errors
- Missing units or final statement
Key Phrases To Include
- Mean = Sum/Number
- Given
- Therefore
The median of five numbers is 15. Four of the numbers are 8, 12, 18, and 20. Find the fifth number.
Marks
3
Topic
Measures of Central Tendency
Difficulty
medium
Template Id
T2
Examiner Tip
Always verify your answer by checking if the median condition is satisfied
Model Answer
Given: Five numbers with median = 15 Four known numbers: 8, 12, 18, 20 To Find: Fifth number Solution: Let the fifth number be x. For median to be 15, when arranged in order, the middle (3rd) number must be 15. Case 1: If x ≤ 15 Arranged order: 8, 12, x, 15, 18, 20 (not possible since we need 15 as median) Case 2: If x > 15 Arranged order: 8, 12, 15, x, 18, 20 (not possible since 15 is not in our original set) Therefore, x = 15 Verification: Arranged order: 8, 12, 15, 18, 20 Median = 15 ✓ Therefore, the fifth number is 15.
Question Type
short_answer
Answer Structure
- Line 1: State given information clearly [0.5 marks]
- Line 2: Define variable for unknown [0.5 marks]
- Line 3: Apply median concept - middle value [1 mark]
- Line 4: Logical reasoning about position [1 mark]
Scoring Breakdown
Marks
1
Criteria
Setting up the problem correctly with variable
Marks
1
Criteria
Understanding that median is the middle value
Marks
1
Criteria
Correct reasoning and final answer
Common Mark Deductions
- Not considering different cases
- Not verifying the answer
- Incorrect understanding of median
Key Phrases To Include
- Let the fifth number be
- middle value
- arranged in order
- verification
A fair coin is tossed 3 times. Find the probability of getting exactly 2 heads.
Marks
3
Topic
Basic Probability
Difficulty
medium
Template Id
T3
Examiner Tip
Always list the sample space systematically - use a tree diagram if needed
Model Answer
Given: Fair coin tossed 3 times To Find: P(exactly 2 heads) Solution: Sample space for 3 tosses: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(S) = 8 Favorable outcomes (exactly 2 heads): E = {HHT, HTH, THH} n(E) = 3 P(exactly 2 heads) = n(E)/n(S) = 3/8 Therefore, the probability of getting exactly 2 heads is 3/8 or 0.375
Question Type
short_answer
Answer Structure
- Line 1: Write given and to find [0.5 marks]
- Line 2: List complete sample space [1 mark]
- Line 3: Identify favorable outcomes [1 mark]
- Line 4: Apply probability formula [0.5 marks]
Scoring Breakdown
Marks
1
Criteria
Correct and complete sample space
Marks
1
Criteria
Correct identification of favorable outcomes
Marks
1
Criteria
Correct probability calculation
Common Mark Deductions
- Incomplete sample space
- Missing favorable outcomes
- Not simplifying fraction
Key Phrases To Include
- Sample space
- favorable outcomes
- P(E) = n(E)/n(S)
Calculate the standard deviation of the data: 2, 4, 6, 8, 10
Marks
5
Topic
Measures of Dispersion
Difficulty
hard
Template Id
T4
Examiner Tip
Show all calculations in a tabular format for clarity and to avoid arithmetic errors
Model Answer
Given: Data set = {2, 4, 6, 8, 10} To Find: Standard deviation (σ) Solution: Step 1: Calculate the mean Mean (x̄) = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6 Step 2: Calculate deviations from mean (xi - x̄): (2-6), (4-6), (6-6), (8-6), (10-6) Deviations: -4, -2, 0, 2, 4 Step 3: Calculate squared deviations (xi - x̄)²: (-4)², (-2)², (0)², (2)², (4)² Squared deviations: 16, 4, 0, 4, 16 Step 4: Calculate variance Variance (σ²) = Σ(xi - x̄)²/n = (16 + 4 + 0 + 4 + 16)/5 = 40/5 = 8 Step 5: Calculate standard deviation Standard deviation (σ) = √variance = √8 = 2√2 = 2.828 Therefore, the standard deviation is 2.828 (or 2√2).
Question Type
long_answer
Answer Structure
- Step 1: Calculate mean [1 mark]
- Step 2: Find deviations from mean [1 mark]
- Step 3: Calculate squared deviations [1 mark]
- Step 4: Calculate variance [1 mark]
- Step 5: Find standard deviation [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct calculation of mean
Marks
1
Criteria
Correct deviations from mean
Marks
1
Criteria
Correct squared deviations
Marks
1
Criteria
Correct variance calculation
Marks
1
Criteria
Correct final standard deviation
Common Mark Deductions
- Calculation errors in any step
- Not showing squared deviations clearly
- Forgetting to take square root for final answer
Key Phrases To Include
- deviations from mean
- squared deviations
- variance
- standard deviation = √variance
What is the mode of the data set: 3, 7, 5, 7, 9, 5, 7, 2?
Marks
1
Topic
Measures of Central Tendency
Difficulty
easy
Template Id
T5
Examiner Tip
Count frequencies systematically and state which value appears most often
Model Answer
Mode = 7 (appears 3 times, most frequent)
Question Type
very_short_answer
Answer Structure
- State the mode value and frequency [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct identification of most frequent value
Common Mark Deductions
- Not identifying the correct frequency
- Giving multiple values without explanation
Key Phrases To Include
- most frequent
- appears
In a class of 30 students, 18 like Mathematics and 12 like Physics. If 8 students like both subjects, find the probability that a randomly selected student likes neither subject.
Marks
3
Topic
Probability with Sets
Difficulty
medium
Template Id
T6
Examiner Tip
Draw a Venn diagram to visualize the problem - it helps avoid double counting
Model Answer
Given: Total students = 30 Like Math = 18, Like Physics = 12 Like both = 8 To Find: P(likes neither subject) Solution: Using the principle of inclusion-exclusion: Students who like at least one subject = Like Math + Like Physics - Like both = 18 + 12 - 8 = 22 Students who like neither subject = Total - Like at least one = 30 - 22 = 8 P(likes neither) = Students who like neither / Total students = 8/30 = 4/15 Therefore, the probability is 4/15 or 0.267
Question Type
short_answer
Answer Structure
- Line 1: State given information [0.5 marks]
- Line 2: Apply inclusion-exclusion principle [1 mark]
- Line 3: Calculate students liking neither [1 mark]
- Line 4: Calculate probability [0.5 marks]
Scoring Breakdown
Marks
1
Criteria
Correct application of inclusion-exclusion principle
Marks
1
Criteria
Correct calculation of students liking neither
Marks
1
Criteria
Correct probability calculation
Common Mark Deductions
- Double counting students
- Not using inclusion-exclusion principle
- Arithmetic errors
Key Phrases To Include
- inclusion-exclusion
- at least one
- neither subject
Find the number of ways to arrange the letters of the word STATISTICS.
Marks
3
Topic
Permutations and Combinations
Difficulty
medium
Template Id
T7
Examiner Tip
List out each letter and its frequency clearly before applying the formula
Model Answer
Given: Word STATISTICS To Find: Number of arrangements Solution: Total letters = 10 Repeated letters: - S appears 3 times - T appears 3 times - A appears 1 time - I appears 2 times - C appears 1 time Number of arrangements = n!/(n₁! × n₂! × n₃! × ...) = 10!/(3! × 3! × 1! × 2! × 1!) = 3,628,800/(6 × 6 × 1 × 2 × 1) = 3,628,800/72 = 50,400 Therefore, there are 50,400 different arrangements.
Question Type
short_answer
Answer Structure
- Line 1: Count total letters and repetitions [1 mark]
- Line 2: Apply permutation formula for repeated objects [1 mark]
- Line 3: Calculate final answer [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct identification of repeated letters
Marks
1
Criteria
Correct application of permutation formula
Marks
1
Criteria
Accurate calculation
Common Mark Deductions
- Missing repeated letters
- Wrong formula
- Calculation errors
Key Phrases To Include
- repeated letters
- n!/(n₁! × n₂!...)
- arrangements
What is the range of the data: 15, 22, 18, 30, 25, 12, 28?
Marks
2
Topic
Measures of Dispersion
Difficulty
easy
Template Id
T8
Examiner Tip
Always scan the data set completely to ensure you find the true maximum and minimum
Model Answer
Given: Data = {15, 22, 18, 30, 25, 12, 28} To Find: Range Solution: Maximum value = 30 Minimum value = 12 Range = Maximum value - Minimum value Range = 30 - 12 = 18 Therefore, the range is 18.
Question Type
short_answer
Answer Structure
- Line 1: Identify maximum and minimum values [1 mark]
- Line 2: Apply range formula and calculate [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct identification of max and min values
Marks
1
Criteria
Correct range calculation
Common Mark Deductions
- Incorrect max/min identification
- Arithmetic error in subtraction
Key Phrases To Include
- maximum value
- minimum value
- Range = Max - Min
A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. Find the probability that both balls are red.
Marks
3
Topic
Conditional Probability
Difficulty
medium
Template Id
T9
Examiner Tip
Remember that 'without replacement' changes the total number of balls for the second draw
Model Answer
Given: 5 red balls, 3 blue balls (total = 8) Two balls drawn without replacement To Find: P(both balls are red) Solution: First draw: P(red) = 5/8 After drawing one red ball: 4 red balls remain out of 7 total Second draw: P(red|first red) = 4/7 P(both red) = P(first red) × P(second red|first red) = (5/8) × (4/7) = 20/56 = 5/14 Therefore, the probability that both balls are red is 5/14.
Question Type
short_answer
Answer Structure
- Line 1: State initial conditions [0.5 marks]
- Line 2: Calculate probability for first draw [0.5 marks]
- Line 3: Calculate conditional probability for second draw [1 mark]
- Line 4: Apply multiplication rule [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct probability for first draw
Marks
1
Criteria
Correct conditional probability for second draw
Marks
1
Criteria
Correct application of multiplication rule
Common Mark Deductions
- Not accounting for replacement
- Wrong conditional probability
- Not simplifying fraction
Key Phrases To Include
- without replacement
- conditional probability
- multiplication rule
In how many ways can 6 people be arranged in a row?
Marks
1
Topic
Permutations and Combinations
Difficulty
easy
Template Id
T10
Examiner Tip
Remember that n! means n × (n-1) × (n-2) × ... × 2 × 1
Model Answer
Number of arrangements = 6! = 720
Question Type
very_short_answer
Answer Structure
- Apply factorial formula and calculate [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct application of 6! = 720
Common Mark Deductions
- Wrong calculation of factorial
- Not using factorial notation
Key Phrases To Include
- 6!
- factorial
The probability of rain on any day is 0.3. Find the probability that it rains on exactly 2 days out of 4 days. (Use binomial distribution)
Marks
5
Topic
Binomial Distribution
Difficulty
hard
Template Id
T11
Examiner Tip
Always identify the distribution type first, then set up the formula with correct parameters
Model Answer
Given: n = 4 days, p = 0.3 (probability of rain), q = 1 - p = 0.7 To Find: P(exactly 2 rainy days) Solution: This follows binomial distribution with parameters n = 4, p = 0.3 For binomial distribution: P(X = r) = ⁿCᵣ × pʳ × q^(n-r) Where: n = 4, r = 2, p = 0.3, q = 0.7 Step 1: Calculate ⁴C₂ ⁴C₂ = 4!/(2! × 2!) = 24/(2 × 2) = 6 Step 2: Calculate p² p² = (0.3)² = 0.09 Step 3: Calculate q^(n-r) q² = (0.7)² = 0.49 Step 4: Apply the formula P(X = 2) = 6 × 0.09 × 0.49 = 6 × 0.0441 = 0.2646 Therefore, the probability is 0.2646 or 26.46%.
Question Type
long_answer
Answer Structure
- Step 1: Identify binomial distribution parameters [1 mark]
- Step 2: Calculate combination ⁴C₂ [1 mark]
- Step 3: Calculate p² and q² [1 mark]
- Step 4: Apply binomial formula [1 mark]
- Step 5: Final calculation and answer [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct identification of binomial distribution
Marks
1
Criteria
Correct calculation of combination
Marks
2
Criteria
Correct application of binomial formula
Marks
1
Criteria
Accurate final calculation
Common Mark Deductions
- Not recognizing binomial distribution
- Errors in combination calculation
- Wrong formula application
Key Phrases To Include
- binomial distribution
- ⁿCᵣ × pʳ × q^(n-r)
- parameters n, p, q
Find the median of the following frequency distribution: Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 5, 8, 12, 10, 5
Marks
4
Topic
Statistics of Grouped Data
Difficulty
hard
Template Id
T12
Examiner Tip
Always create the cumulative frequency table systematically and double-check your median class identification
Model Answer
Given: Frequency distribution table To Find: Median Solution: Step 1: Prepare cumulative frequency table Class | Frequency | Cumulative Frequency 0-10 | 5 | 5 10-20 | 8 | 13 20-30 | 12 | 25 30-40 | 10 | 35 40-50 | 5 | 40 Total frequency (N) = 40 Step 2: Find median position Median position = N/2 = 40/2 = 20th observation Step 3: Identify median class The 20th observation lies in class 20-30 (since cf = 25 > 20) Step 4: Apply median formula Median = l + [(N/2 - cf)/f] × h Where: l = 20, N/2 = 20, cf = 13, f = 12, h = 10 Median = 20 + [(20 - 13)/12] × 10 = 20 + (7/12) × 10 = 20 + 5.83 = 25.83 Therefore, the median is 25.83.
Question Type
long_answer
Answer Structure
- Step 1: Create cumulative frequency table [1 mark]
- Step 2: Find N/2 position [1 mark]
- Step 3: Identify median class [1 mark]
- Step 4: Apply median formula [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct cumulative frequency calculation
Marks
1
Criteria
Correct median position N/2
Marks
1
Criteria
Correct median class identification
Marks
1
Criteria
Correct application of median formula
Common Mark Deductions
- Errors in cumulative frequency
- Wrong median class
- Formula application errors
Key Phrases To Include
- cumulative frequency
- median class
- Median = l + [(N/2 - cf)/f] × h
Two dice are thrown simultaneously. What is the probability of getting a sum of 8?
Marks
2
Topic
Basic Probability
Difficulty
easy
Template Id
T13
Examiner Tip
List all favorable outcomes systematically - consider (a,b) and (b,a) as different outcomes
Model Answer
Given: Two dice thrown simultaneously To Find: P(sum = 8) Solution: Total possible outcomes = 6 × 6 = 36 Favorable outcomes for sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) Number of favorable outcomes = 5 P(sum = 8) = Favorable outcomes/Total outcomes = 5/36 Therefore, the probability is 5/36.
Question Type
short_answer
Answer Structure
- Line 1: Calculate total possible outcomes [0.5 marks]
- Line 2: List favorable outcomes [1 mark]
- Line 3: Apply probability formula [0.5 marks]
Scoring Breakdown
Marks
1
Criteria
Correct enumeration of favorable outcomes
Marks
1
Criteria
Correct probability calculation
Common Mark Deductions
- Missing some favorable outcomes
- Wrong total outcomes
- Not listing outcomes systematically
Key Phrases To Include
- total outcomes = 36
- favorable outcomes
- ordered pairs
In how many ways can a committee of 4 people be selected from 10 people?
Marks
2
Topic
Permutations and Combinations
Difficulty
medium
Template Id
T14
Examiner Tip
Remember: combinations for selection (order doesn't matter), permutations for arrangement (order matters)
Model Answer
Given: Total people = 10, Committee size = 4 To Find: Number of ways to select committee Solution: This is a combination problem (order doesn't matter) Number of ways = ¹⁰C₄ = 10!/(4! × 6!) = (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) = 5040/24 = 210 Therefore, there are 210 ways to select the committee.
Question Type
short_answer
Answer Structure
- Line 1: Identify as combination problem [0.5 marks]
- Line 2: Apply combination formula [1 mark]
- Line 3: Calculate final answer [0.5 marks]
Scoring Breakdown
Marks
1
Criteria
Correct identification and formula setup
Marks
1
Criteria
Accurate calculation
Common Mark Deductions
- Using permutation instead of combination
- Calculation errors
- Wrong formula
Key Phrases To Include
- combination
- ¹⁰C₄
- order doesn't matter
The heights of students in a class are normally distributed with mean 160 cm and standard deviation 5 cm. Find the probability that a randomly selected student has height between 155 cm and 165 cm.
Marks
4
Topic
Normal Distribution
Difficulty
hard
Template Id
T15
Examiner Tip
Always convert to standard normal first, then use the symmetry property: Φ(-z) = 1 - Φ(z)
Model Answer
Given: μ = 160 cm, σ = 5 cm Heights are normally distributed To Find: P(155 < X < 165) Solution: Step 1: Standardize the values using Z = (X - μ)/σ For X = 155: Z₁ = (155 - 160)/5 = -5/5 = -1 For X = 165: Z₂ = (165 - 160)/5 = 5/5 = 1 Step 2: Find the required probability P(155 < X < 165) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) = Φ(1) - Φ(-1) Step 3: Use standard normal table Φ(1) = 0.8413 Φ(-1) = 0.1587 Step 4: Calculate final probability P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 Therefore, the probability is 0.6826 or 68.26%.
Question Type
long_answer
Answer Structure
- Step 1: Convert to standard normal using Z-scores [1 mark]
- Step 2: Set up probability expression [1 mark]
- Step 3: Use standard normal table values [1 mark]
- Step 4: Calculate final answer [1 mark]
Scoring Breakdown
Marks
1
Criteria
Correct Z-score calculations
Marks
1
Criteria
Correct probability setup
Marks
1
Criteria
Correct use of normal table
Marks
1
Criteria
Accurate final calculation
Common Mark Deductions
- Incorrect Z-score calculation
- Wrong probability setup
- Table reading errors
Key Phrases To Include
- standardize
- Z-score
- standard normal table
- Φ(z)
Mark Wise Strategy
Dos
- Write the final answer clearly
- Use correct mathematical notation
- Include units if applicable
Donts
- Waste time on lengthy explanations
- Show unnecessary working steps
- Leave answer unclear
Marks
1
Strategy
Direct application of formula or concept with minimal working
Expected Length
1 line with final answer
Time Allocation
30 seconds to 1 minute
Dos
- State the formula used
- Show key substitution step
- Box the final answer
- Include proper mathematical notation
Donts
- Skip the formula statement
- Make arithmetic errors
- Forget units in final answer
Marks
2
Strategy
Show formula, substitution, and calculation clearly
Expected Length
3-4 lines showing key steps
Time Allocation
2-3 minutes
Dos
- Write 'Given' and 'To Find'
- Show all major steps
- Include intermediate calculations
- Verify answer if time permits
Donts
- Jump steps without explanation
- Make calculation errors
- Forget to state conclusion
Marks
3
Strategy
Show complete method with reasoning and verification where possible
Expected Length
5-7 lines with logical flow
Time Allocation
3-4 minutes
Dos
- Break into numbered steps
- Show all formula derivations
- Include diagrams where helpful
- Verify final answer
- State assumptions if any
Donts
- Rush through steps
- Skip verification
- Make multiple arithmetic errors
- Use unclear notation
Marks
5
Strategy
Complete step-by-step solution with clear reasoning at each stage
Expected Length
8-12 lines with detailed working
Time Allocation
6-8 minutes
General Answer Writing Tips
- Always write 'Given:' and 'To Find:' clearly at the start of numerical problems
- Show all formula substitutions step-by-step - never skip intermediate steps
- Box or underline your final answer clearly with appropriate units
- Draw neat diagrams for probability tree diagrams and Venn diagrams
- Define statistical terms before using them in your answers
- Use mathematical symbols correctly (σ for standard deviation, μ for mean)
- Always verify your probability answers are between 0 and 1
- Round final answers to appropriate decimal places as specified in the question
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