CUET UG Applied Mathematics Booster Test 2 - Probability Distribution and Expectation -

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CUET UG Applied Mathematics Booster Test 2 - Probability Distribution and Expectation

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Probability Distribution and Expectation

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QUESTION 1 OF 20

A random variable X has the following probability distribution: P(X=-1) = 0.1, P(X=0) = 0.8, P(X=1) = k, P(X=2) = 0.05. For this to be a valid distribution table, what is the value of k?

Answer Review

1 A random variable X has the following probability distribution: P(X=-1) = 0.1, P(X=0) = 0.8, P(X=1) = k, P(X=2) = 0.05. For this to be a valid distribution table, what is the value of k?

A 0.001

B 0.05

C 0.1

D 0.2

💡 Short Explanation

�� Sum of probabilities must equal 1 �� Known sum = 0.95 �� Remaining probability gives k = 0.05

✅ Correct Answer Explanation

For a valid probability distribution, total probability must satisfy ΣP(X)=1 → 0.1 + 0.8 + k + 0.05 = 1 → 0.95 + k = 1 ⇒ k = 0.05 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Total probability becomes less than 1
�� Option C → Total exceeds 1
�� Option D → Total exceeds 1

🎯 Strategy

Used

Elimination

Application: Use normalization condition ΣP(X)=1

Final Logic: Balance remaining probability to satisfy unity

🧠 Memory Trick

"Probabilities always complete to 1"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Probability Distribution –

2 Match the random variable X (number of heads in 3 coin tosses) with the correct probability outcome mapping.

List I	List II
(A) X = 0	(I) 3/8
(B) X = 1	(II) 1/8
(C) X = 2	(III) 1/8
(D) X = 3	(IV) 3/8

A (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

C (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

D (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

💡 Short Explanation

�� Binomial distribution n=3, p=1/2 �� Probabilities follow 1/8, 3/8, 3/8, 1/8 �� Match accordingly

✅ Correct Answer Explanation

Total outcomes = 2³ = 8 → P(X=0)=1/8, P(X=1)=3/8, P(X=2)=3/8, P(X=3)=1/8 → Matching with given list gives correct mapping in option A

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignment of binomial probabilities
�� Option C → Reverses correct symmetry pattern
�� Option D → Completely mismatches distribution structure

🎯 Strategy

Used

Option Grouping

Application: Use binomial symmetry (1-3-3-1 pattern)

Final Logic: Match standard coin toss distribution

🧠 Memory Trick

"3 toss → 1,3,3,1 pattern"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Distribution –

3 In a two-dice rolling experiment, which of the following probabilities for the sum of digits X are mathematically correct?

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

�� Total outcomes = 36 �� Check standard sum frequencies �� Only 12 case is incorrect

✅ Correct Answer Explanation

Sum 2 → (1,1) → 1/36 correct → Sum 5 → 4 outcomes → 4/36 correct → Sum 7 → 6 outcomes → 6/36 correct → Sum 12 → only (6,6) → 1/36, so 2/36 is wrong → Hence A, B, C are correct

❓ Why Other Options Are Incorrect

�� Option A → Excludes correct B and C
�� Option B → Misses correct C
�� Option D → Incorrect probability for sum 12

🎯 Strategy

Used

Elimination

Application: Use dice outcome frequency table

Final Logic: Reject incorrect extreme-case probability

🧠 Memory Trick

"7 is most frequent, extremes least"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Two Dice Experiment –

4 Identify the INCORRECT statement regarding the distribution of the sum of two dice (X).

A The distribution is symmetrical, peaking at X = 7.

B P(X = 3) and P(X = 11) have exactly the same probability value.

C The sum of all outcome probabilities equals 1.

D The mathematical expectation E(X) for the sum of two dice is 6.

💡 Short Explanation

�� Mean of two dice is 7 �� Distribution is symmetric �� Expectation statement is incorrect

✅ Correct Answer Explanation

E(single die)=3.5 → E(two dice)=3.5+3.5=7 → So expectation is 7, not 6 → Hence option D is incorrect

❓ Why Other Options Are Incorrect

�� Option A → True symmetry property
�� Option B → True equal probability pairs
�� Option C → Valid probability rule

🎯 Strategy

Used

Conceptual Recall

Application: Use expectation of uniform discrete variable

Final Logic: Correct mean is 7

🧠 Memory Trick

"Two dice average = 7"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Expectation of Random Variables –

5 A coin is tossed 3 times. What is the mean expectation E(X)?

A 1.0

B 1.5

C 2.0

D 3.0

💡 Short Explanation

�� Binomial distribution �� n = 3, p = 1/2 �� Mean = np

✅ Correct Answer Explanation

E(X)=np → = 3 × 1/2 = 1.5 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates mean
�� Option C → Overestimates mean
�� Option D → Maximum value, not expectation

🎯 Strategy

Used

Substitution

Application: Apply binomial expectation formula

Final Logic: Multiply n and p

🧠 Memory Trick

"Mean = n × p"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Mean –

6 In a biased coin toss experiment, P(Head) = 2/3 and P(Tail) = 1/3. If the coin is tossed twice, what is P(X=1)?

A 2/9

B 4/9

C 5/9

D 8/9

💡 Short Explanation

�� Exactly one head in two trials �� Two arrangements possible �� Apply binomial formula

✅ Correct Answer Explanation

P(X=1)=C(2,1)(2/3)(1/3) → =2 × 2/3 × 1/3 = 4/9 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates probability
�� Option C → Not derived correctly
�� Option D → Exceeds valid probability logic

🎯 Strategy

Used

Substitution

Application: Use binomial probability formula

Final Logic: Count arrangements × probability

🧠 Memory Trick

"One success in two tosses = 2 ways"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Distribution –

7 A variable X takes values 1, 2, 3, 4, 5 with probabilities 0.1, 0.4, 0.05, -0.2, 0.2. Why does this NOT qualify as a valid probability distribution?

A The values of X cannot be greater than 1.

B Probabilities cannot be negative numbers.

C The sum of probabilities exceeds 1.

D Discrete variables cannot have exactly 5 outcomes.

💡 Short Explanation

�� Probability must be ≥ 0 �� One value is negative �� Distribution becomes invalid

✅ Correct Answer Explanation

Valid probability requires 0 ≤ P(X) ≤ 1 → Here P(4)= -0.2 violates rule → Hence distribution is invalid

❓ Why Other Options Are Incorrect

�� Option A → X can take any real discrete values
�� Option C → Sum condition is not the issue
�� Option D → Number of outcomes is valid

🎯 Strategy

Used

Elimination

Application: Check probability axioms

Final Logic: Negative probability invalidates model

🧠 Memory Trick

"No negative probability allowed"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Valid Distribution Rules –

8 For a variable Y taking values -1, 0, 4, 5 with probabilities 1/7, 2/7, 3/7, 1/7. What ensures these are valid probabilities before calculating E(Y)?

A All p_i are positive fractions and their sum equals exactly 1.

B The values of Y are strictly positive.

C The highest probability maps to the median.

D E(Y) evaluates to zero.

💡 Short Explanation

�� Check probability axioms �� Non-negative values �� Sum equals 1

✅ Correct Answer Explanation

Valid probability distribution requires: → P(i) ≥ 0 and ΣP(i)=1 → 1/7+2/7+3/7+1/7=1 → Hence valid distribution

❓ Why Other Options Are Incorrect

�� Option B → Values of Y need not be positive
�� Option C → No such rule in probability
�� Option D → Expectation is not validity condition

🎯 Strategy

Used

Substitution

Application: Verify normalization condition

Final Logic: Probability axioms satisfied

🧠 Memory Trick

"Valid = positive + sum 1"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Probability Axioms –

9 If a player wins Rs 5 for rolling an even digit and loses Rs 2 for an odd digit, what is E(X)?

A Rs 1.5

B Rs 3.0

C Rs 1.0

D Rs 0.0

💡 Short Explanation

�� Even probability = 1/2 �� Odd probability = 1/2 �� Weighted expectation

✅ Correct Answer Explanation

E(X)=5(1/2)+(-2)(1/2) → = (5−2)/2 = 1.5 → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Overestimates gain
�� Option C → Incorrect weighting
�� Option D → Incorrect neutrality assumption

🎯 Strategy

Used

Weighted Average

Application: Compute expected gain-loss balance

Final Logic: Net expectation is 1.5

🧠 Memory Trick

"Gain minus loss, then halve"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Mathematical Expectation –

10 In a distribution, the outcome vector is X = (0, 1, 2) and probability vector is P = (1/4, 1/2, 1/4). The weighted average E(X) is:

A 0.5

B 0.75

C 1.0

D 1.5

💡 Short Explanation

�� Weighted sum of outcomes �� Multiply and add probabilities �� Final result = 1

✅ Correct Answer Explanation

E(X)=0(1/4)+1(1/2)+2(1/4) → = 0 + 0.5 + 0.5 = 1 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates mean
�� Option B → Partial sum
�� Option D → Overestimates result

🎯 Strategy

Used

Substitution

Application: Direct weighted mean computation

Final Logic: Sum of weighted values equals 1

🧠 Memory Trick

"Multiply, then add = expectation"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Expected Value –

11 A class of 20 students has marks showing a distribution. If Σ (xᵢ * pᵢ) evaluates to 385 / 20, what is the mean expectation E(X)?

A 19.25

B 14.50

C 21.00

D 13.90

💡 Short Explanation

�� Expectation equals weighted sum �� Convert fraction to decimal �� Final mean = 19.25

✅ Correct Answer Explanation

E(X) = Σ(xᵢ pᵢ) → = 385 / 20 → = 19.25 → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Incorrect scaling
�� Option C → Not derived from given fraction
�� Option D → Miscalculated division

🎯 Strategy

Used

Substitution

Application: Direct evaluation of expectation formula

Final Logic: Divide numerator by denominator

🧠 Memory Trick

"Expectation = sum of products"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Mathematical Expectation –

12 The mathematical variance Var(X) is essentially derived from expectations. Which formula accurately connects these products?

A Var(X) = [E(X)]² - E(X²)

B Var(X) = E(X²) - [E(X)]²

C Var(X) = E(X) - E(X²)

D Var(X) = Σ(xᵢ - pᵢ)²

💡 Short Explanation

�� Variance measures spread �� Based on expectation of squares �� Standard identity formula

✅ Correct Answer Explanation

Variance is defined as: → Var(X) = E(X²) − [E(X)]² → Measures deviation from mean → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Sign reversed (incorrect identity)
�� Option C → Dimensionally incorrect
�� Option D → Incorrect formulation (mixes probability and values wrongly)

🎯 Strategy

Used

Conceptual Recall

Application: Use standard variance identity

Final Logic: Correct formula is E(X²) − [E(X)]²

🧠 Memory Trick

"Square mean minus mean square? No—reverse it!"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Variance of Random Variable –

📄 Passage

PASSAGE MCQs (COMPULSORY) Have a look at the probability distribution of two different random variables X and Y. For X: Values (1, 2, 3, 4) have probabilities (1/7, 3/7, 2/7, 1/7). The mean expectation E(X) evaluates to 17/7. For Y: Values (-1, 0, 4, 5) have probabilities (1/7, 2/7, 3/7, 1/7). The mean expectation E(Y) calculates to (-1/7 + 0 + 12/7 + 5/7) = 16/7. (Note: Source text has a minor anomaly stating they have equal means, but actual calculation shows 17/7 vs 16/7. Assuming the principle holds that different spreads can yield comparable or identical means). In such cases, we need a technique to check variability and the extent to which the values of the random variable are spread out from the mean.

Based on the passage, what specific statistical tool is needed to differentiate between distributions that might have identical or similar means but different data spreads?

A Standard Z-Score

B Variance

C Binomial limit

D Moving Average

💡 Short Explanation

�� Mean alone is insufficient �� Spread comparison needed �� Variance measures dispersion

✅ Correct Answer Explanation

Variance measures how far values spread from mean → Two distributions can have same mean but different variance → Hence variance is the correct tool

❓ Why Other Options Are Incorrect

�� Option A → Standardizes values, not spread comparison tool
�� Option C → Not related to spread measurement
�� Option D → Smooths data, not dispersion analysis

🎯 Strategy

Used

Contextual Matching

Application: Identify tool for variability comparison

Final Logic: Variance directly measures dispersion

🧠 Memory Trick

"Same mean → compare variance"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Measures of Dispersion –

📄 Passage

Based on the passage calculations, what is the exact product sum for the Y variable's expectation E(Y)?

A 15/7

B 16/7

C 17/7

D 1

💡 Short Explanation

�� Compute weighted sum �� Add products correctly �� Final expectation given

✅ Correct Answer Explanation

E(Y) = Σ xᵢ pᵢ → = (-1)(1/7) + (0)(2/7) + (4)(3/7) + (5)(1/7) → = -1/7 + 0 + 12/7 + 5/7 → = 16/7 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect summation
�� Option C → Confuses X and Y distributions
�� Option D → Does not match computed value

🎯 Strategy

Used

Substitution

Application: Direct weighted sum calculation

Final Logic: Add all weighted contributions

🧠 Memory Trick

"Multiply then add carefully"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Expectation Calculation –

15 If a discrete distribution features X = {-2, 2, 5} with probabilities P(X) = {0.5, 0.2, 0.3}, what is the theoretical mean E(X)?

A 0.5

B 0.9

C 1.5

D 2.0

💡 Short Explanation

�� Multiply values with probabilities �� Sum all terms �� Compute expectation

✅ Correct Answer Explanation

E(X)=(-2)(0.5)+(2)(0.2)+(5)(0.3) → = -1 + 0.4 + 1.5 → = 0.9 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates result
�� Option C → Overestimates sum
�� Option D → Incorrect computation

🎯 Strategy

Used

Substitution

Application: Weighted sum computation

Final Logic: Add all contributions correctly

🧠 Memory Trick

"Negative first, then balance positives"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Mean of Random Variable –

16 The Greek letter μ (mu) is universally utilized to represent which specific parameter of a probability distribution?

A Variance

B Standard Deviation

C Mean Expectation E(X)

D Total Probability

💡 Short Explanation

�� μ denotes population mean �� Central tendency measure �� Expected value of distribution

✅ Correct Answer Explanation

In probability theory, μ represents mean (expected value) → It measures central location of distribution → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Variance is denoted by σ²
�� Option B → Standard deviation is σ
�� Option D → Total probability is always 1, not μ

🎯 Strategy

Used

Conceptual Recall

Application: Match symbols with statistical meaning

Final Logic: μ corresponds to mean

🧠 Memory Trick

"μ = mean"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Statistics & Probability – Topic: Measures of Central Tendency –

17 If two variables X and Y have equal means but Var(X) = 2.4 and Var(Y) = 8.1, what does this indicate about their distributions?

A Y's sample elements are clustered closer to the mean than X's.

B X's sample elements have a wider range than Y's.

C Y's sample elements are spread farther from the mean than X's.

D Both distributions are perfectly identical.

💡 Short Explanation

�� Higher variance → more spread �� Compare 2.4 vs 8.1 �� Y has greater dispersion

✅ Correct Answer Explanation

Variance measures spread from mean → 8.1 > 2.4, so Y is more dispersed → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Opposite of variance meaning
�� Option B → Incorrect comparison direction
�� Option D → Variance differs, so not identical

🎯 Strategy

Used

Comparison Analysis

Application: Compare magnitude of variance values

Final Logic: Higher variance means more spread

🧠 Memory Trick

"Bigger variance = wider spread"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Comparison of Distributions –

18 The variance Var(X) measures the average degree to which each sample element differs from the:

A Highest outcome value

B Mean of the sample space

C Lowest probability

D Standard normal curve

💡 Short Explanation

�� Variance measures deviation �� Always around mean �� Not extremes or probabilities

✅ Correct Answer Explanation

Variance is defined as average squared deviation from mean → It measures spread around expected value → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Variance not based on maximum value
�� Option C → Probability unrelated to deviation
�� Option D → Standard normal is a distribution, not reference point

🎯 Strategy

Used

Conceptual Recall

Application: Identify definition of variance

Final Logic: Deviation is measured from mean

🧠 Memory Trick

"Variance = distance from mean"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Variance Definition –

19 In the television viewing model, if P(X=0)=0.2, P(X=1)=0.16, P(X=2)=0.32, P(X=3)=0.32, what is the probability that the person watches AT LEAST 2 hours of television?

A 0.32

B 0.48

C 0.64

D 0.80

💡 Short Explanation

�� Add probabilities for X ≥ 2 �� Combine X=2 and X=3 �� Final sum = 0.64

✅ Correct Answer Explanation

P(X≥2)=P(2)+P(3) → =0.32+0.32 → =0.64 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Only one component
�� Option B → Partial addition
�� Option D → Overestimates total

🎯 Strategy

Used

Option Grouping

Application: Sum required event probabilities

Final Logic: Add relevant outcomes only

🧠 Memory Trick

"≥ means add all above"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Event Probability –

20 In the same television model, what is the calculated expectation E(X) representing the average daily hours watched?

A 1.50

B 1.76

C 2.00

D 2.24

💡 Short Explanation

�� Weighted sum of outcomes �� Multiply and add probabilities �� Compute expectation

✅ Correct Answer Explanation

E(X)=0(0.2)+1(0.16)+2(0.32)+3(0.32) → = 0 + 0.16 + 0.64 + 0.96 → = 1.76 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates total
�� Option C → Incorrect aggregation
�� Option D → Overestimates sum

🎯 Strategy

Used

Substitution

Application: Direct expectation calculation

Final Logic: Sum weighted outcomes

🧠 Memory Trick

"Multiply each, then add all"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Expected Value –

CUET UG Applied Mathematics Booster Test 2 - Probability Distribution and Expectation

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