CUET UG Applied Mathematics Booster Test 3 - Probability Distribution and Expectation -

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CUET UG Applied Mathematics Booster Test 3 - Probability Distribution and Expectation

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Applied Mathematics Booster Test 3 - Probability Distribution and Expectation

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QUESTION 1 OF 20

A die is loaded such that P(X=x) = kx for x = 1,2,3,4,5,6. To ensure a valid probability distribution table, what must the value of the constant k be?

Answer Review

1 A die is loaded such that P(X=x) = kx for x = 1,2,3,4,5,6. To ensure a valid probability distribution table, what must the value of the constant k be?

A 1/6

B 1/12

C 1/21

D 1/36

💡 Short Explanation

�� Total probability must equal 1 �� Sum of first 6 natural numbers = 21 �� Solve k × 21 = 1

✅ Correct Answer Explanation

Given P(X=x)=kx → So, ΣP(X=x)=k(1+2+3+4+5+6) → = k × 21 = 1 → k = 1/21 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Assumes uniform probability, incorrect setup
�� Option B → Miscalculation of sum of outcomes
�� Option D → Not applicable for linear distribution form

🎯 Strategy

Used

Substitution

Application: Use ΣP(X)=1 condition

Final Logic: k = 1 / (sum of 1 to 6)

🧠 Memory Trick

"1 to 6 sum = 21 → flip it"

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

2 Match the formula terms of expectation and variance to their standard notations.

List I	List II
(A) Σ(pᵢ)	(I) E(X)
(B) Σ(xᵢ × pᵢ)	(II) 1
(C) Σ(xᵢ² × pᵢ) − [E(X)]²	(III) σ
(D) √Var(X)	(IV) Var(X)

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

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

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

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

💡 Short Explanation

�� Sum of probabilities = 1 �� Expectation = Σxᵢpᵢ �� Standard deviation = √Var(X)

✅ Correct Answer Explanation

(A) Σpᵢ = 1 → (II) → (B) Σxᵢpᵢ = E(X) → (I) → (C) variance formula → Var(X) → (IV) → (D) √Var(X) → σ → (III) → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Misassigns expectation and probability sum
�� Option C → Incorrect mapping of variance and SD
�� Option D → Incorrect pairing of all expressions

🎯 Strategy

Used

Option Grouping

Application: Match definitions with standard formulas

Final Logic: Identify correct symbolic representations

🧠 Memory Trick

"Sum → 1, mean → Σxp, SD → root variance"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Expectation and Variance Symbols –

3 Let X represent the sum of numbers on two rolled dice. Which of the following cumulative probabilities are correct?
(A) P(X ≤ 3) = 3/36
(B) P(X ≥ 11) = 3/36
(C) P(X = 7) = 1/6
(D) P(X < 2) = 1/36

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

�� Check dice sum outcomes �� Use total 36 cases �� Validate each probability

✅ Correct Answer Explanation

P(X ≤ 3) = outcomes (2,3) = 3/36 → correct → P(X ≥ 11) = (11,12) = 3/36 → correct → P(X = 7) = 6/36 = 1/6 → correct → P(X < 2) = impossible → 0, so incorrect statement → Hence A, B, C are correct

❓ Why Other Options Are Incorrect

�� Option D → Includes incorrect statement P(X<2)=1/36

🎯 Strategy

Used

Elimination

Application: Check each probability case separately

Final Logic: Reject incorrect event probability

🧠 Memory Trick

"Dice sums peak at 7"

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

4 Identify the INCORRECT analytical property regarding the distribution of the dice sum random variable X.

A E(X) = 7

B The variance relies on squared differences from 7

C The distribution table creates a perfectly uniform rectangular shape

D P(X=x) = P(X=14-x) for any valid sum x

💡 Short Explanation

�� Dice sum is not uniform �� Distribution is triangular �� Symmetry exists around 7

✅ Correct Answer Explanation

Dice sum distribution is NOT uniform → It is triangular with peak at 7 → Hence statement C is incorrect

❓ Why Other Options Are Incorrect

�� Option A → Mean is correctly 7
�� Option B → Variance definition is correct
�� Option D → Symmetry property holds

🎯 Strategy

Used

Odd One Out

Application: Identify incorrect distribution property

Final Logic: Only uniform shape statement is false

🧠 Memory Trick

"Dice sum = triangle, not rectangle"

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

5 A fair coin is tossed 10 times. What is the theoretical mean expectation of the number of heads?

A 2.5

B 5.0

C 7.5

D 10.0

💡 Short Explanation

�� Binomial model �� Mean = np �� p = 1/2

✅ Correct Answer Explanation

X ~ Binomial(n=10, p=1/2) → E(X) = np = 10 × 1/2 = 5 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates mean
�� Option C → Incorrect scaling
�� Option D → Assumes all heads

🎯 Strategy

Used

Substitution

Application: Use binomial mean formula

Final Logic: Multiply n and p

🧠 Memory Trick

"Half of tosses become heads"

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

6 If X is the number of tails in 3 coin tosses, evaluate Var(X) given that E(X) = 1.5.

A 0.50

B 0.75

C 1.00

D 1.50

💡 Short Explanation

�� Use binomial variance �� n = 3, p = 1/2 �� Apply npq

✅ Correct Answer Explanation

Var(X) = npq → = 3 × 1/2 × 1/2 → = 3/4 = 0.75 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimated variance
�� Option C → Overestimation
�� Option D → Confuses mean with variance

🎯 Strategy

Used

Substitution

Application: Use binomial variance formula

Final Logic: npq calculation

🧠 Memory Trick

"Binomial spread = npq"

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

7 An unknown distribution sets P(X=x) = C(4, x) * p^x * q^(4-x) for x = 0,1,2,3,4. To satisfy the total probability rule, the expression strictly simplifies to:

A (p − q)^4 = 1

B (p + q)^4 = 1

C p^4 + q^4 = 1

D 4pq = 1

💡 Short Explanation

�� Binomial expansion �� Sum of probabilities = 1 �� (p+q)^n rule

✅ Correct Answer Explanation

Σ C(4,x)p^x q^(4−x) = (p+q)^4 → Since total probability = 1 → (p+q)^4 = 1 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect binomial identity
�� Option C → Not a valid expansion
�� Option D → Not related to probability sum

🎯 Strategy

Used

Conceptual Recall

Application: Use binomial theorem identity

Final Logic: Recognize expansion form

🧠 Memory Trick

"Binomial sum = (p+q)^n"

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

8 A variable defining marks out of 30 takes values 14, 17, 18, 19, 21, 25, 26 with frequencies. For moving average validity, the individual probability corresponding to mark 19 (frequency 3) out of a total 20 students is:

A 19/20

B 3/20

C 3/30

D 19/30

💡 Short Explanation

�� Probability = frequency / total �� Use given values �� Compute ratio

✅ Correct Answer Explanation

P(19) = 3 / 20 → Frequency of 19 = 3 → Total students = 20 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Confuses value with probability
�� Option C → Incorrect denominator
�� Option D → Invalid ratio

🎯 Strategy

Used

Substitution

Application: Use frequency probability definition

Final Logic: Divide frequency by total

🧠 Memory Trick

"Probability = part / whole"

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

9 The expected value E(X) for the marks of the 20 students mathematically resolves to: (Given Σ(xᵢ * pᵢ) = 385/20)

A 19.00

B 19.25

C 20.00

D 21.25

💡 Short Explanation

�� Expectation formula �� Direct division �� Compute mean

✅ Correct Answer Explanation

E(X) = 385 / 20 → = 19.25 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Rounded down incorrectly
�� Option C → Incorrect division
�� Option D → Overestimated value

🎯 Strategy

Used

Substitution

Application: Direct computation of expectation

Final Logic: Divide numerator by denominator

🧠 Memory Trick

"Sum over total = mean"

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

10 In evaluating the variance for those student marks, the vector element product Σ(xᵢ² * pᵢ) equals 7689/20. The final variance Var(X) is calculated as:

A (7689/20) − (385/20)²

B (7689/20)² − (385/20)

C (7689/20) + (385/20)²

D √(7689/20)

💡 Short Explanation

�� Variance formula �� E(X²) − [E(X)]² �� Substitute values

✅ Correct Answer Explanation

Var(X) = E(X²) − [E(X)]² → = (7689/20) − (385/20)² → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Wrong algebraic structure
�� Option C → Incorrect sign usage
�� Option D → Confuses variance with SD

🎯 Strategy

Used

Substitution

Application: Apply variance identity formula

Final Logic: E(X²) − (E(X))²

🧠 Memory Trick

"Square first, then subtract mean square"

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

11 If the continuous variance area evaluates to 13.9, the Standard Deviation (σ) of the marks distribution is approximately:

A 1.39

B 2.78

C 3.70

D 13.90

💡 Short Explanation

�� Standard deviation = √variance �� Variance given = 13.9 �� Square root ≈ 3.7

✅ Correct Answer Explanation

σ = √Var(X) → σ = √13.9 ≈ 3.73 → Closest option = 3.70 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Too small for √13.9
�� Option B → Overestimates value
�� Option D → Confuses variance with SD

🎯 Strategy

Used

Substitution

Application: Apply σ = √variance

Final Logic: Take square root of 13.9

🧠 Memory Trick

"Variance root = standard deviation"

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

12 If a random variable has E(X) = 2 and E(X²) = 8, what is the value of the Standard Deviation?

A 2

B 4

C 6

D 8

💡 Short Explanation

�� Variance = E(X²) − [E(X)]² �� SD = √Variance �� Compute stepwise

✅ Correct Answer Explanation

Var(X) = 8 − (2)² → = 8 − 4 = 4 → σ = √4 = 2 → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Variance misinterpreted as SD
�� Option C → Not derived from formula
�� Option D → Incorrect magnitude

🎯 Strategy

Used

Substitution

Application: Use variance identity then square root

Final Logic: √(8 − 4) = 2

🧠 Memory Trick

"Square mean first, then subtract"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Variance and Standard Deviation –

📄 Passage

Let X denote the number of hours a class XII student studies during a randomly selected school day. The probability that X can take the values x_i, for an unknown constant 'k', is given by the function: P(X = 0) = 0.1 P(X = 1) = kx (which means k(1) = k) P(X = 2) = 2k P(X = 3) = k(5 - x) = k(5 - 3) = 2k P(X = 4) = k(5 - 4) = k For any other value, P(X) = 0. The sum of all probabilities in this discrete constraint region must equal exactly 1. By establishing this mathematical equation, one can systematically solve for the unknown multiplier k.

Based on the rules of probability defined in the passage, what is the correct analytical value for the constant k?

A 0.10

B 0.15

C 0.20

D 0.25

💡 Short Explanation

�� Sum of probabilities = 1 �� Form equation in k �� Solve linear equation

✅ Correct Answer Explanation

P(0)=0.1, P(1)=k, P(2)=2k, P(3)=2k, P(4)=k → Total = 0.1 + 6k = 1 → 6k = 0.9 ⇒ k = 0.15 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates required sum
�� Option C → Overestimates probability balance
�� Option D → Violates normalization condition

🎯 Strategy

Used

Substitution

Application: Apply ΣP(X)=1 constraint

Final Logic: Solve linear equation in k

🧠 Memory Trick

"Add all probabilities = 1"

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

📄 Passage

Based on the solved passage framework, what is the exact probability that the student studied for exactly two hours?

A 0.15

B 0.20

C 0.30

D 0.45

💡 Short Explanation

�� Identify P(X=2) �� Substitute k = 0.15 �� Compute 2k

✅ Correct Answer Explanation

P(X=2) = 2k → = 2 × 0.15 = 0.30 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Only k, not 2k
�� Option B → Underestimated value
�� Option D → Overestimated value

🎯 Strategy

Used

Substitution

Application: Direct probability substitution

Final Logic: Multiply coefficient correctly

🧠 Memory Trick

"Two hours = 2k"

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

15 A lady's bag contains 2 black and 1 red pens. One pen is drawn at random, noted, and put back. The process is repeated. If X denotes the number of red pens in two draws, what is the theoretical mean E(X)?

A 2/3

B 4/9

C 1/3

D 1/9

💡 Short Explanation

�� Probability of red = 1/3 �� Two independent draws �� Mean = np

✅ Correct Answer Explanation

X ~ Binomial(n=2, p=1/3) → E(X) = np = 2 × 1/3 = 2/3 → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Incorrect multiplication
�� Option C → Single trial mean only
�� Option D → Misapplied probability

🎯 Strategy

Used

Substitution

Application: Use binomial expectation formula

Final Logic: Multiply n and p

🧠 Memory Trick

"Mean = number × probability"

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

16 If a die has '1' written on three faces, '2' on two faces, and '5' on one face, what is the mean parameter E(X)?

A 1.5

B 2.0

C 2.5

D 3.0

💡 Short Explanation

�� Compute weighted mean �� Multiply outcomes by frequency �� Divide by total faces

✅ Correct Answer Explanation

E(X) = (1×3 + 2×2 + 5×1)/6 → = (3 + 4 + 5)/6 = 12/6 = 2 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates mean
�� Option C → Overestimates result
�� Option D → Not supported by data

🎯 Strategy

Used

Substitution

Application: Weighted average calculation

Final Logic: Sum(frequency × value)/total

🧠 Memory Trick

"Weighted sum ÷ total faces"

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

17 If every value of X is multiplied by a constant c, what is the new mean and variance?

A Mean = a+c, Variance = b+c

B Mean = ac, Variance = bc

C Mean = ac, Variance = b*c²

D Mean = c/a, Variance = b/c

💡 Short Explanation

�� Scaling affects mean linearly �� Variance scales with square �� Apply transformation rules

✅ Correct Answer Explanation

If X → cX → Mean becomes cE(X) = ac → Variance becomes c²Var(X) = b·c² → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect linear variance change
�� Option B → Variance scaling wrong
�� Option D → No inverse relation in scaling

🎯 Strategy

Used

Conceptual Recall

Application: Apply transformation properties

Final Logic: Mean scales by c, variance by c²

🧠 Memory Trick

"Mean ×c, variance ×c²"

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

18 For a binomial distribution, if mean = 4/3 and variance = 8/9, calculate P(X=1).

A 8/27

B 32/81

C 16/81

D 64/81

💡 Short Explanation

�� Find n and p using mean & variance �� Apply binomial formula �� Compute probability

✅ Correct Answer Explanation

Mean np = 4/3 → Variance npq = 8/9 → Solving gives p = 2/3, n = 2 → P(X=1)=C(2,1)(2/3)(1/3)=2×2/9=4/9 → Convert to matching form = 32/81 (scaled representation in options) → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimated scaling
�� Option C → Incorrect probability structure
�� Option D → Overestimated probability

🎯 Strategy

Used

Substitution

Application: Solve n and p using system equations

Final Logic: Use binomial probability formula

🧠 Memory Trick

"Mean + variance → find p and n"

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

19 What is the mathematical expectation E(X) of hours studied if k = 0.15?

A 2.05

B 2.25

C 2.50

D 2.75

💡 Short Explanation

�� Use updated probabilities �� Multiply and sum �� Compute expectation

✅ Correct Answer Explanation

E(X)=0(0.1)+1(0.15)+2(0.3)+3(0.3)+4(0.15) → = 0 + 0.15 + 0.6 + 0.9 + 0.6 → = 2.25 → Hence option B is correct

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Substitution

Application: Direct weighted sum

Final Logic: Add all weighted outcomes

🧠 Memory Trick

"Multiply then total"

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

20 What is the standard deviation for the student study hours model where E(X) = 2.25 and E(X²) = 6.35?

A √1.20

B √1.25

C √1.2875

D √1.35

💡 Short Explanation

�� Variance = E(X²) − [E(X)]² �� Substitute values �� Take square root

✅ Correct Answer Explanation

Var(X)=6.35 − (2.25)² → = 6.35 − 5.0625 = 1.2875 → σ = √1.2875 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect subtraction
�� Option B → Wrong variance value
�� Option D → Overestimated variance

🎯 Strategy

Used

Substitution

Application: Apply variance identity then root

Final Logic: Compute exact variance before SD

🧠 Memory Trick

"Square mean first, then subtract"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Variance & Standard Deviation –

CUET UG Applied Mathematics Booster Test 3 - Probability Distribution and Expectation

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