CUET UG Applied Mathematics Booster Test 3 - Variance and Binomial Distribution -

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CUET UG Applied Mathematics Booster Test 3 - Variance and Binomial Distribution

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Variance and Binomial Distribution

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QUESTION 1 OF 20

Let X be a discrete random variable taking values -1, 0, and 1 with probabilities 0.2, 0.5, and 0.3 respectively. What is the precise calculated value of the variance Var(X)?

Answer Review

1 Let X be a discrete random variable taking values -1, 0, and 1 with probabilities 0.2, 0.5, and 0.3 respectively. What is the precise calculated value of the variance Var(X)?

A 0.1

B 0.49

C 0.50

D 0.69

💡 Short Explanation

The experiment consists of drawing a sweater, replacing it, and drawing a sweater again (sampling with replacement).

The two black sweaters are distinct individuals (B_1,B_2) and the white sweater is W_1, giving 3 choices for each of the two draws.

The total number of outcomes in the sample space is 3×3=9 ordered pairs.

✅ Correct Answer Explanation

To compute the variance of a discrete random variable, we use the standard algebraic formula Var(X)=E(X^2)-[E(X)]^2. First, we find the mathematical expectation (mean), E(X): E(X)=(-1×0.2)+(0×0.5)+(1×0.3)=-0.2+0+0.3=0.1 Second, we calculate the expected value of the squared variable, E(X^2): E(X^2)=([-1]^2×0.2)+([0]^2×0.5)+([1]^2×0.3)=(1×0.2)+0+(1×0.3)=0.2+0.3=0.5 Third, substitute these parameters back into the variance identity: Var(X)=E(X^2)-[E(X)]^2=0.5-(0.1)^2=0.5-0.01=0.49 This matches Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 0.1 is the mean E(X), not the variance Var(X).
Option C → Incorrect because 0.50 is E(X^2), failing to subtract the square of the mean ([E(X)]^2).
Option D → Incorrect because 0.69 is a numerical miscalculation that does not match the probability weights.

🎯 Strategy

Used: Direct Quantification

Application: Apply the foundational discrete distribution expectations step-by-step.

Final Logic: Computing the components yields 0.5-0.01=0.49, which confirms Option B.

🧠 Memory Trick

Variance is the "mean of squares" minus the "square of the mean." Don't forget to subtract 0.1^2 from 0.5.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Mean and Variance of a Random Variable

2 Match the specific intermediate calculation steps to their resulting values for the student marks distribution dataset (mean = 19.25).

List I	List II
A. ∑(xᵢ · pᵢ)	I 13.90
B. ∑(xᵢ² · pᵢ)	II 19.25
C. [E(X)]²	III 384.45
D. Var(X)	IV 370.5625

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

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

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

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

💡 Short Explanation

A random variable is mathematically defined as a real-valued function defined on a sample space.

The domain of this function is the sample space (S) of the random experiment.

The co-domain is the entire set of real numbers (R), while the range contains the actual mapped values.

✅ Correct Answer Explanation

Let's map each structural mathematical term in List I to its calculated value in List II: (A) ∑_^((x_i⋅p_i)): This formula defines the mathematical expectation E(X) (the population mean). The question states the mean is 19.25, so (A) maps to (II). (C) [E(X)]^2: This is the square of the expectation calculated above. (19.25)^2=370.5625, so (C) maps to (IV). (B) ∑_^((x_i^2⋅p_i)): This represents E(X^2), the mean of the squared data points. For this dataset, it equals 384.45, so (B) maps to (III). (D) Var(X): Calculated as E(X^2)-[E(X)]^2=384.45-370.5625=13.90, so (D) maps to (I). Combining these pairs gives (A)-(II), (B)-(III), (C)-(IV), (D)-(I), matching Option A.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it maps the mean ∑_^((x_i⋅p_i)) to 13.90, which is the variance value.
Option C → Incorrect because it matches the basic mean expression with the squared term value 384.45.
Option D → Incorrect because it assigns the square of the mean value 370.5625 directly to the simple expectation expression.

🎯 Strategy

Used: Elimination

Application: Start by pairing the mean expression with its given numeric value (A→II).

Final Logic: Knowing that A pairs with II eliminates options B, C, and D, pointing directly to Option A.

🧠 Memory Trick

∑_^(x_i)p_i is always the mean. Since the text tells you the mean is 19.25, A must match II.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Components of Variance Formula

3 Which of the following analytical identities correctly outline the calculation of variance and standard deviation from expected parameters?
(A) Var(X)=E(X^2)-(E(X))^2
(B) σ=√(Var(X))
(C) E(X^2)=Var(X)+(E(X))^2
(D) Var(X)=E(X)-[E(X^2)]^2
Options format:

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

A random variable maps outcomes from a sample space to real numbers.
The formal definition requires it to be a real-valued function, meaning imaginary outputs are excluded.
Statements A, B, and C are correct, while statement D is mathematically false.

✅ Correct Answer Explanation

Let's evaluate each identity based on statistical theory: (A) Var(X)=E(X^2)-(E(X))^2: This is the fundamental, mathematically verified formula for computing variance, so it is correct. (B) σ=√(Var(X)): By definition, the standard deviation is the positive square root of the variance, so it is correct. (C) E(X^2)=Var(X)+(E(X))^2: Rearranging formula (A) by moving (E(X))^2 to the other side gives this exact equation, making it algebraically correct. (D) Var(X)=E(X)-[E(X^2)]^2: This formulation mixes up the terms and squares the wrong component, making it incorrect. Since statements A, B, and C are correct, Option C is the correct choice.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it includes statement D, which is a mathematically invalid rearrangement of the terms.
Option B → Incorrect because it leaves out statement C, which is a completely valid algebraic manipulation of the variance formula.
Option D → Incorrect because it includes all options, failing to recognize that statement D is false.

🎯 Strategy

Used: Elimination

Application: Evaluate statement D first to see if it is a valid variance formula.

Final Logic: Since statement D is incorrect, any option containing D (Options A and D) is eliminated, which leads to Option C.

🧠 Memory Trick

Standard deviation is always the square root of variance (B), and variance is the mean of squares minus the square of the mean (A). Equation C is just moving terms around.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Algebraic Identities of Variance

4 Identify the INCORRECT statement concerning the properties of Variance.

A Variance can never be a negative value because it is an expected value of squared deviations.

B If Var(X)=0, it means the random variable X is a constant (no spread).

C Var(X)=∑_^((x_i^2⋅p_i))-(∑_^((x_i⋅p_i)))^2.

D The standard deviation is simply the variance divided by two.

💡 Short Explanation

The co-domain of a random variable is always the entire set of real numbers (R).

A random variable can take on negative values, zero, fractional values, or decimals depending on the experiment.

Restricting the co-domain strictly to positive integers is incorrect, making Option B the correct choice for this question.

✅ Correct Answer Explanation

Let's review the mathematical properties of variance to find the false statement: Option A is true: Because variance averages squared distances (∑_^((x_i-μ)^2)p_i), and squares cannot be negative, variance is always ≥0. Option B is true: A variance of 0 means there is absolutely no deviation or spread around the mean, which only happens if the random variable is a constant value. Option C is true: This is the standard operational formula for discrete distributions, written using summation symbols. Option D is false: The standard deviation is the positive square root of the variance (σ=√(Var(X))). Dividing the variance by 2 is a completely different operation. Therefore, Option D is the correct choice for this incorrect statement question.

❓ Why Other Options Are Incorrect

Option A → This is a true statement; variance cannot be negative because it is built from squared terms.
Option B → This is a true statement; zero variance means there is no variation, which defines a constant.
Option C → This is a true statement; it correctly expresses the standard variance formula using summation format.

🎯 Strategy

Used: Keyword Filter

Application: Look for mathematical relationship words like "divided by two" and compare them with standard formulas.

Final Logic: Since standard deviation relies on a square root rather than division, Option D is identified as the incorrect statement.

🧠 Memory Trick

Standard deviation uses a square root (√(Var)), not division (Var/2). Don't mix up roots with halves.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Properties of Variance and Standard Deviation

5 For a given distribution, if E(X)=2.67 and E(X^2)=8.0, what is the approximate standard deviation (mixture spread limit) of this distribution?

A 0.89

B 0.94

C 1.63

D 5.33

💡 Short Explanation

The input sample outcome is HT (Head followed by Tail).

The random variable X counts the total number of heads present within that specific outcome.

Inspecting the outcome HT shows it contains exactly one head, so X(HT)=1.

✅ Correct Answer Explanation

To find the standard deviation, we first calculate the variance using the parameters provided: Var(X)=E(X^2)-[E(X)]^2 Substitute E(X^2)=8.0 and E(X)=2.67 into the formula: Var(X)=8.0-(2.67)^2=8.0-7.1289=0.8711 The standard deviation (σ) is the positive square root of the variance: σ=√(0.8711)≈0.9333 Rounding to two decimal places gives approximately 0.94, which matches Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 0.89 is close to the raw variance value (0.87), forgetting to take the final square root.
Option C → Incorrect because 1.63 is a numerical error that does not follow the square root calculation of these parameters.
Option D → Incorrect because 5.33 is obtained by multiplying 2.67 by 2 instead of following the correct formula steps.

🎯 Strategy

Used: Direct Quantification

Application: Run the numbers through the two-step formula: calculate the variance first, then take its square root.

Final Logic: The math yields √(0.8711)≈0.94, confirming Option B as the correct answer.

🧠 Memory Trick

Always follow the two steps: Step 1 is subtracting (8.0-2.67^2=0.87). Step 2 is taking the square root (√(0.87)≈0.94).

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Standard Deviation Calculations

6 In a binomial constraint region, the variance is given by npq. Knowing that p+q=1, what is the absolute maximum value the variance Var(X) can take for a fixed number of trials n?

A n

B n/2

C n/4

D n/8

💡 Short Explanation

The sample space contains four distinct outcomes: {HH,HT,TH,TT}.

The functional mapping rule for Y is: "Number of Tails"-"Number of Heads" .

Evaluating all outcomes gives the values -2,0, and 2, which forms the range of the function.

✅ Correct Answer Explanation

In a binomial distribution, the variance function is defined as V(p)=npq. Since q=1-p, we can rewrite the variance as a function of a single variable: V(p)=np(1-p)=np-np^2 To find where this expression reaches its maximum, we take its derivative with respect to p and set it to zero: dV/dp=n-2np=0⟹2np=n⟹p=1/2 When p=1/2, the failure probability is q=1-1/2=1/2. Substituting these values back into the variance formula gives the maximum possible variance: Var_(max)=n(1/2)(1/2)=n/4 This mathematically matches Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because n would require pq=1, which is impossible since the maximum value of the product p(1-p) is 0.25.
Option B → Incorrect because n/2 overestimates the upper bound of the product of two probabilities that sum to 1.
Option D → Incorrect because n/8 is smaller than the true maximum value produced when success and failure chances are equal.

🎯 Strategy

Used: Direct Formula Application

Application: Use calculus optimization or the arithmetic mean-geometric mean (AM-GM) inequality on the product pq.

Final Logic: The product pq reaches its maximum value of 1/4 when p=q=0.5, making the maximum variance n/4 (Option C).

🧠 Memory Trick

A binomial product pq is largest when things are perfectly split (0.5×0.5=0.25=1/4). So the maximum variance is always n/4.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Mathematical Properties of Binomial Variance

7 (Expected Marks Index) In the dataset of 20 students' marks where E(X^2)=7689/20 and [E(X)]^2=(385/20)^2, the variance computation involves fractions. What does (7689/20)-(148225/400) exactly resolve to?

A 5555 / 400

B 13.9

C Both A and B are correct

D 3.7

💡 Short Explanation

Rolling a standard six-sided die results in outcomes from the sample space {1,2,3,4,5,6}.

The input value is 6, which is an even number.

The rule states that rolling an even number results in winning Rs 5, so X(6)=5.

✅ Correct Answer Explanation

Let's evaluate the fraction subtraction step-by-step to check the options: The given expression is: 7689/20-148225/400 To subtract these, convert 7689/20 to have the common denominator of 400 by multiplying the numerator and denominator by 20: 7689×20/20×20=153780/400 Now, subtract the numerators over the common denominator: 153780-148225/400=5555/400 This matches the fraction given in Option A. Next, convert this fraction into its decimal form: 5555/400=13.8875 Rounding this value to one decimal place gives 13.9, which matches Option B. Since both the exact fraction form (A) and the rounded decimal form (B) are correct representations, Option C ("Both A and B are correct") is the correct choice.

❓ Why Other Options Are Incorrect

Option A → This is correct, but choosing it alone misses the equally valid decimal representation described in option B.
Option B → This is correct, but choosing it alone overlooks the exact fraction form described in option A.
Option D → Incorrect because 3.7 is a wrong value that does not equal the result of the fraction subtraction.

🎯 Strategy

Used: Direct Quantification

Application: Calculate the exact common fraction and its decimal form to check all choices.

Final Logic: Since the calculation matches both A and B, Option C is the correct answer.

🧠 Memory Trick

Always look at all options before deciding. If a fraction and its decimal form are both listed, the answer is usually "Both."

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Fraction Handling in Variance Calculations

8 If the moving average (mean) of a binomial distribution is 4/3 and its variance is 8/9, what is the calculated number of finite trials 'n'?

A 2

B 3

C 4

D 6

💡 Short Explanation

The range of a random variable is the set of all actual numerical values it can output.

For any odd number rolled, the function outputs a value of -2.

For any even number rolled, the function outputs a value of +5, making the final range {-2,+5}.

✅ Correct Answer Explanation

In a binomial distribution, the mean and variance are determined by the parameters n (number of trials) and p (probability of success). We are given: 1. Mean=np=4/3 2. Variance=npq=8/9 To find the probability of failure (q), divide the variance equation by the mean equation: q=npq/np=8/9/4/3=8/9×3/4=24/36=2/3 Since p+q=1, we can find the success probability p: p=1-q=1-2/3=1/3 Now substitute the value of p back into the mean equation (np=4/3): n(1/3)=4/3⟹n=4 This confirms that the number of finite trials is 4, matching Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because if n=2 with p=1/3, the mean would be 2/3, which contradicts the given mean of 4/3.
Option B → Incorrect because if n=3 with p=1/3, the mean would be 1, which does not match the problem parameters.
Option D → Incorrect because if n=6, it violates the algebraic relationship established by dividing the given variance by the mean.

🎯 Strategy

Used: Direct Formula Application

Application: Divide the variance by the mean to find q, compute p, and then solve for n.

Final Logic: The calculations show that n=4 is the only value that satisfies both equations, pointing to Option C.

🧠 Memory Trick

Divide variance by mean to get q (8/9/4/3=2/3). This means p=1/3. Since n×1/3=4/3, n must be 4.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Finding Binomial Parameters

9 Rohit shoots a target with a success probability of p = 1/6. To find the minimum number of chances (n) required so the probability of hitting the target at least once is more than 0.99, which inequality correctly models the failure constraint?

A 1-(1/6)^n>0.99

B (5/6)^n≤0.01

C (1/6)^n≤0.01

D 1-(5/6)^n<0.99

💡 Short Explanation

Random variables are classified based on the characteristics of the values they can take on.
A variable that takes on isolated, distinct points on the number line is classified as discrete.
Examples include counts of items, where fractional values between consecutive outcomes do not exist.

✅ Correct Answer Explanation

The problem asks for the target hitting probability to be strictly greater than 0.99: P(at least one hit)>0.99 Using the complement rule, we can rewrite "at least one hit" as 1-P(all failures). Since each shot is independent, the probability of failing all n times is q^n. Given p=1/6, the failure probability is q=1-1/6=5/6. Substitute this into the probability inequality: 1-(5/6)^n>0.99 Now, rearrange the inequality to isolate the failure term. Subtract 0.99 from both sides and move the fractional term to the right side: 1-0.99>(5/6)^n⟹0.01>(5/6)^n Rewriting this with the fractional term on the left side flips the inequality sign: (5/6)^n<0.01 This constraint matches the boundary model represented in Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it uses the success probability 1/6 inside the failure term, misapplying the complement rule.
Option C → Incorrect because it uses the success probability 1/6 instead of the failure probability 5/6 to model total failure.
Option D → Incorrect because the inequality sign is flipped the wrong way, stating that the success probability is less than 0.99.

🎯 Strategy

Used: Direct Formula Application

Application: Set up the complement probability rule 1-q^n>0.99 and solve the inequality.

Final Logic: Rearranging the equation yields (5/6)^n<0.01, confirming Option B as the correct model.

🧠 Memory Trick

At least once means 1-Failure^n. Moving things around gives Failure^n<0.01. Since failure is 5/6, look for (5/6)^n≤0.01.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Complementary Probability in Independent Trials

10 Sonal and Anannya throw a die alternately to get a '1'. Because the independent trials alternate, Sonal winning involves calculating an infinite geometric sequence vector. If her probability of winning on her first turn is 1/6, and she throws on turns 1, 3, 5..., what is the common ratio 'r' of her winning probability series?

A 1/6

B 5/6

C 25/36

D 5/36

💡 Short Explanation

For a valid probability distribution, individual probabilities must be non-negative (p_i≥0).

The sum of all probabilities across the entire sample space must equal exactly 1 (∑▒p_i =1).

This condition satisfies the fundamental normalization axiom of probability theory.

✅ Correct Answer Explanation

Let's analyze the game sequence. Sonal throws first (turn 1), Anannya throws second (turn 2), Sonal throws third (turn 3), and so on. For anyone rolling the die, the probability of rolling a '1' (Success, S) is 1/6, and the probability of rolling any other number (Failure, F) is 5/6. Sonal can win the game during her specific turns: 1. Turn 1: She rolls a success immediately ⟹P(Win on Turn 1)=p=1/6 2. Turn 3: She fails on turn 1, Anannya fails on turn 2, and Sonal succeeds on turn 3 ⟹P(Win on Turn 3)=F×F×S=q×q×p=q^2p=(5/6)^2(1/6) 3. Turn 5: Everyone fails on turns 1, 2, 3, and 4, and Sonal succeeds on turn 5 ⟹P(Win on Turn 5)=q^4p=(5/6)^4(1/6) Writing this out as an infinite series gives: Series=p+q^2p+q^4p+…=1/6+(25/36)1/6+(25/36)^21/6+… An examining of this geometric progression shows that each term is multiplied by a common ratio r=q^2. Calculating this value gives: r=(5/6)^2=25/36 This matches Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 1/6 is the probability of success on a single roll, not the ratio between alternating turns.
Option B → Incorrect because 5/6 is the single-turn failure probability, which would be the ratio if they were tracking consecutive turns instead of alternating turns.
Option D → Incorrect because 5/36 is a calculation error that does not represent the squared failure probability.

🎯 Strategy

Used: Direct Formula Application

Application: Expand the probability tracking series to isolate the multiplier between consecutive terms.

Final Logic: The multiplier between Sonal's turns is two consecutive failures (q^2), which evaluates to 25/36 (Option C).

🧠 Memory Trick

Sonal skips a turn every time Anannya plays, so the ratio must account for two failures in a row (q×q). 5/6×5/6=25/36.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Infinite Geometric Series in Alternating Games

11 To fulfill the finite trials condition for binomial expansion, if a player tosses a coin 10 times, the total probability area evaluates to (1/2 + 1/2)^10. What is the value of the 5th term in this expansion representing exactly 4 successes?

A C(10,5) * (1/2)^10

B C(10,4) * (1/2)^10

C C(10,6) * (1/2)^10

D C(10,4) * (1/2)^4

💡 Short Explanation

A continuous probability distribution is represented by a probability density function (PDF).
The total area bounded under the curve over the entire domain represents the absolute certainty of all possible outcomes.
In accordance with the fundamental axioms of probability, the sum of all possibilities must always equal 1 (or 100%).

✅ Correct Answer Explanation

According to the binomial theorem, the expansion of (q+p)^n is written as a sum of terms where the general (r+1)-th term is defined as: T_(r+1)=(n/r)⋅q^(n-r)⋅p^r To find the 5th term (T_5), we set r=4. Given that the total number of coin tosses is n=10, and the coin is fair (p=1/2,q=1/2), we substitute these parameters into the formula: T_5=(10/4)⋅(1/2)^(10-4)⋅(1/2)^4 T_5=(10/4)⋅(1/2)^6⋅(1/2)^4 Using exponents rules, we add the powers together: (1/2)^6⋅(1/2)^4=(1/2)^(10). This gives the final simplified form: T_5=(10/4)⋅(1/2)^(10) This matches Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it sets r=5, which calculates the 6th term (representing 5 successes) instead of the 5th term.
Option C → Incorrect because it uses a combination index of 6, which calculates the 7th term in the binomial expansion.
Option D → Incorrect because it truncates the total exponent power to 4, failing to account for the accompanying failure probabilities.

🎯 Strategy

Used: Direct Formula Application

Application: Apply the general term formula T_(r+1) with r=4 and n=10.

Final Logic: Combining the base fractions simplifies the exponent to 10, pointing directly to Option B.

🧠 Memory Trick

The 5th term has r=4. The combination must start with C(10,4), and since p=q=1/2, the overall exponent is always the total trials (10).

📚 Class XI – Mathematics – Chapter: Binomial Theorem / Class XII – Probability: Binomial Distribution

12 A continuous transition from Binomial to Poisson distribution involves the integral limit where the constant success probability p becomes very small while n approaches infinity. In this condition, the variance npq approaches the value of:

A Zero

B Infinity

C The Mean (np)

D One

💡 Short Explanation

For continuous random variables, summing up individual probabilities is done using definite integration.

The function f(x) represents a valid probability density function (PDF).

Integrating f(x) across its entire possible range from -∞ to ∞ calculates the total probability, which equals 1.

✅ Correct Answer Explanation

The Poisson distribution is used to model rare events by taking the limit of a binomial distribution as the number of trials (n) grows very large and the success probability (p) becomes very small, keeping their product (np=λ) stable. The binomial variance is calculated as Var(X)=npq. Since q=1-p, we can look at the limit as p approaches 0: lim_(p→0)q=lim_(p→0)(1-p)=1 Substituting this limiting value into the variance formula gives: Var(X)=np⋅q→np⋅(1)=np Since np represents the mean of the distribution, the variance approaches the value of the mean (λ). A defining characteristic of the Poisson distribution is that its mean and variance are exactly equal, confirming Option C as the correct answer.

❓ Why Other Options Are Incorrect

Option A → Incorrect because while p→0, the number of trials n→∞ prevents the overall variance product from dropping to zero.
Option B → Incorrect because the product np is held stable at a constant rate λ, preventing the variance from growing to infinity.
Option D → Incorrect because the variance stabilizes at the specific parameter value λ (np), not necessarily at the number one.

🎯 Strategy

Used: Direct Formula Application

Application: Evaluate the mathematical limit of the variance formula np(1-p) as p→0.

Final Logic: Since (1-p)→1, the expression simplifies directly to np, which represents the distribution's mean (Option C).

🧠 Memory Trick

In a Poisson distribution, the mean and the variance are twin properties—they are exactly equal (Mean=Variance=np).

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Poisson Limit of Binomial Parameters

📄 Passage

The Binomial expansion (q + p)^n is the mathematical engine driving the binomial distribution table. When computing the probability of 'r' successes, P(r) = C(n,r) * p^r * q^(n-r), we are essentially extracting the (r + 1)th term of this expansion. For example, if a fair coin is tossed 9 times, calculating the probability of getting at least 5 tails involves summing the probabilities of exactly 5, 6, 7, 8, and 9 tails. This relies on the sum: C(9,5)p⁵q⁴ + C(9,6)p⁶q³ + C(9,7)p⁷q² + C(9,8)p⁸q¹ + C(9,9)p⁹q⁰. Because the coin is fair, p = q = 1/2, meaning the p and q terms can be factored out completely as (1/2)⁹.

Based on the binomial expansion logic in the passage, what does the summation [C(9,5) + C(9,6) + C(9,7) + C(9,8) + C(9,9)] analytically evaluate to using combinations symmetry?

A 256

B 126

C 512

D 63

💡 Short Explanation

The passage states that an hour/minute watch skips directly from minute to minute with no intermediate steps shown.
The passage explicitly characterizes this specific time change as "distinct and countable."
This step-by-step progression matches the core definition of a discrete variable given in the text.

✅ Correct Answer Explanation

According to the properties of binomial coefficients, the sum of all coefficients for a given power n equals 2^n. For n=9 tosses, the total sum of all combinations is: (9/0)+(9/1)+…+(9/9)=2^9=512 By the symmetry property of combinations, (n/r)=(n/n-r). This means the 10 terms in this expansion can be paired into two identical groups: [(9/0)+(9/1)+(9/2)+(9/3)+(9/4)]=[(9/5)+(9/6)+(9/7)+(9/8)+(9/9)] Since these two groups are equal and add up to 512, each group must equal exactly half of the total: Target Sum=512/2=256 This matches Option A.

❓ Why Other Options Are Incorrect

Option B → Incorrect because 126 is the value of the single coefficient (9/4) or (9/5), rather than the sum of the entire upper half.
Option C → Incorrect because 512 represents the total sum of all 10 coefficients combined, forgetting to divide by two using symmetry.
Option D → Incorrect because 63 does not follow the base-2 exponential growth rules of binomial coefficients.

🎯 Strategy

Used: Direct Formula Application

Application: Apply binomial coefficient summation laws and symmetry properties to split the total sum.

Final Logic: Dividing the total coefficient sum (2^9=512) by 2 yields 256, confirming Option A.

🧠 Memory Trick

The total sum of all combinations is 2^9=512. Because of symmetry, the upper half must be exactly half of that total: 512/2=256.

📚 Class XI – Mathematics – Chapter: Binomial Theorem – Topic: Properties of Coefficients

📄 Passage

Based on the passage setup, what is the final calculated probability of getting at least 5 tails in 9 tosses of a fair coin?

A 126 / 512

B 256 / 512

C 1 / 2

D Both B and C are correct

💡 Short Explanation

The text contrasts the hour/minute watch with a watch that tracks continuous elapsed seconds.
According to the final sentence, a continuous random variable is defined by its capacity to take on a set of infinite and uncountable values.
This description directly matches the explanation provided in the passage.

✅ Correct Answer Explanation

We calculate the probability by following the steps outlined in the passage. For a fair coin where p=q=1/2, the probability of getting at least 5 tails is written as: P(X≥5)=[(9/5)+(9/6)+(9/7)+(9/8)+(9/9)]×(1/2)^9 From our calculation in Question 13, the sum of these combinations is 256. We also know that (1/2)^9=1/512. Multiplying these together gives: P(X≥5)=256/512 This matches Option B. Next, simplify this fraction by dividing the numerator and the denominator by 256: P(X≥5)=256÷256/512÷256=1/2 This matches Option C. Since both the raw fraction (B) and the simplified fraction (C) are correct descriptions of the final probability, Option D ("Both B and C are correct") is the correct choice.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 126 is the value of a single coefficient component, leading to an incorrect probability value.
Option B → This fraction is correct, but choosing it alone overlooks the simplified form (1/2) described in option C.
Option C → This simplified form is correct, but choosing it alone misses the raw fraction representation described in option B.

🎯 Strategy

Used: Direct Quantification

Application: Calculate the exact probability fraction and simplify it to verify all options.

Final Logic: Since the calculation yields both 256/512 and 1/2, Option D is the correct choice.

🧠 Memory Trick

Tossing a fair coin an odd number of times (9) means the probability of getting "at least half" is always exactly a 50-50 choice (1/2 or 256/512).

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Binomial Distribution Probability Calculations

15 A binomial distribution has a mean of 4 and a variance of 3. What is the required number of trials (n) to construct this distribution?

A 7

B 12

C 16

D 20

💡 Short Explanation

A sample space contains the raw outcomes of an experiment, such as coin tosses.
Multiple different functions can be defined on these identical raw outcomes to track different metrics.
For example, one variable can count the number of heads, while another tracks the number of tails or consecutive matches.

✅ Correct Answer Explanation

In a binomial distribution, the mean and variance are linked by the parameters n and p. We are given: 1. Mean=np=4 2. Variance=npq=3 To find the probability of failure (q), divide the variance by the mean: q=npq/np=3/4 Since the total probability of an event must equal 1 (p+q=1), we find the success probability p: p=1-q=1-3/4=1/4 Now, substitute this value of p back into the mean equation (np=4) to find the total number of trials: n×(1/4)=4⟹n=4×4=16 This confirms that the distribution requires exactly 16 trials, matching Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 7 trials would produce fractional parameters that do not match a mean of 4.
Option B → Incorrect because if n=12 with a mean of 4, the success probability would be p=4/12=1/3, which gives a variance of 12×1/3×2/3=8/3≠3.
Option D → Incorrect because 20 trials does not satisfy the algebraic relationship between the given mean and variance.

🎯 Strategy

Used: Direct Formula Application

Application: Divide variance by mean to calculate q, find p, and solve for n.

Final Logic: Running the parameters through the formulas shows that n=16 is the only valid answer, pointing to Option C.

🧠 Memory Trick

Variance / Mean gives q=3/4, so p=1/4. If a quarter of the trials equals 4, then the total number of trials must be 16.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Computing Binomial Trial Parameters

16 In the probability table for B(4, 1/3), what is the strict ratio of the probability of exactly 2 successes to the probability of exactly 1 success? (P(X=2) / P(X=1))

A 1/2

B 2/3

C 3/4

D 1/1

💡 Short Explanation

A probability distribution provides a complete summary of an experiment's outcomes and their likelihoods.
It systematically pairs each distinct value of a random variable with its calculated probability.
This layout ensures that the entire sample space is clearly mapped and accounts for a total probability of 1.

✅ Correct Answer Explanation

For a binomial distribution B(n,p) with parameters n=4 and p=1/3, the probability of failure is q=1-1/3=2/3. We find the probabilities using the binomial formula P(X=r)=(n/r)p^rq^(n-r): 1. For exactly 1 success (r=1): P(X=1)=(4/1)(1/3)^1(2/3)^3=4×1/3×8/27=32/81 1. For exactly 2 successes (r=2): P(X=2)=(4/2)(1/3)^2(2/3)^2=6×1/9×4/16 false, it's 4/9=6×4/81=24/81 Now, find the ratio of P(X=2) to P(X=1): Ratio=P(X=2)/P(X=1)=24/81/32/81=24/32=3/4 This matches Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 1/2 is the ratio of p to q, not the final ratio of the two probabilities.
Option B → Incorrect because 2/3 does not match the simplified fraction of the calculated success terms.
Option D → Incorrect because 1/1 assumes that P(X=1) and P(X=2) are exactly equal, which is disproved by calculating the terms.

🎯 Strategy

Used: Direct Quantification

Application: Expand both probability terms using the binomial formula and simplify their ratio.

Final Logic: The ratio simplifies to 24/32=3/4, making Option C the correct choice.

🧠 Memory Trick

Write out the ratio formula: Ratio/=n-r+1/r×p/q. For r=2, this gives 4-2+1/2×1/3/2/3=3/2×1/2=3/4.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Ratios of Consecutive Binomial Probabilities

17 If a coin is biased such that the probability of getting a head is three times the probability of getting a tail. If the coin is tossed 8 times, what is the theoretical mean number of heads?

A 2

B 4

C 6

D 8

💡 Short Explanation

Standard probability problems assume the use of a fair, unbiased coin.

A fair coin means that heads and tails share an identical chance of occurring (1/2 each).

For a two-coin toss, this fair baseline ensures that all four outcomes (HH,HT,TH,TT) have an equal probability of 1/4.

✅ Correct Answer Explanation

First, we determine the individual probabilities for the biased coin. Let the probability of getting a tail be q. The problem states that the probability of getting a head (p) is three times that value: p=3q Since getting a head or a tail are the only two possible outcomes, their probabilities must sum to 1: p+q=1⟹3q+q=1⟹4q=1⟹q=1/4 Substituting q=1/4 back into the equation gives the success probability (getting a head): p=3×(1/4)=3/4 The theoretical mean (μ) of a binomial distribution is found by multiplying the total number of trials (n=8) by the success probability (p): Mean=np=8×3/4=2×3=6 This matches Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 2 is the expected number of tails (nq=8×1/4=2), not heads.
Option B → Incorrect because 4 assumes a fair coin (p=0.5), which ignores the bias described in the problem.
Option D → Incorrect because 8 would mean getting a head is guaranteed on every single toss (p=1), which is false.

🎯 Strategy

Used: Direct Formula Application

Application: Set up the probability equation 3q+q=1 to find p, then apply the mean formula μ=np.

Final Logic: Evaluating the parameters gives 8×3/4=6, confirming Option C.

🧠 Memory Trick

Out of 4 parts, 3 are heads. Since the coin is tossed 8 times, heads will show up on 3/4 of the tosses: 8×3/4=6.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Mean of a Biased Binomial Distribution

18 For the same biased coin (p = 3/4, q = 1/4) tossed 8 times, what is the calculated variance?

A 1.0

B 1.5

C 2.0

D 6.0

💡 Short Explanation

A trial that results in one of exactly two possible outcomes is named after mathematician Jacob Bernoulli.
These two outcomes are conventionally classified as "Success" and "Failure."
Each trial must be independent, meaning the outcome of one trial does not affect the next.

✅ Correct Answer Explanation

To find the variance of a binomial distribution, we use the formula Var(X)=npq, where n is the number of trials, p is the probability of success, and q is the probability of failure. From the previous problem, we have: n=8 trials p=3/4 (probability of getting a head) q=1/4 (probability of getting a tail) Substitute these values directly into the variance formula: Var(X)=8×3/4×1/4=8×3×1/4×4=24/16 Simplifying this fraction by dividing the numerator and denominator by 8 gives: Var(X)=3/2=1.5 This matches Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 1.0 is a round number that does not match the product of these fractional parameters.
Option C → Incorrect because 2.0 is the mean number of tails, not the variance of the distribution.
Option D → Incorrect because 6.0 is the calculated mean (np) of the distribution, forgetting to multiply by the failure probability q.

🎯 Strategy

Used: Direct Formula Application

Application: Substitute the parameters n=8,p=3/4,q=1/4 into the variance equation npq.

Final Logic: Multiplying the components gives exactly 1.5, pointing directly to Option B.

🧠 Memory Trick

Variance is just the Mean multiplied by q. Since the mean is 6 and q=1/4, the variance is 6×1/4=1.5.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Variance of a Binomial Distribution

19 A fair coin is tossed 10 times. What is the probability of getting at most 6 heads?

A 193 / 512

B 53 / 64

C 105 / 512

D 848 / 1024

💡 Short Explanation

Rolling two standard six-sided dice yields a total of 6×6=36 equally likely outcomes.

The random variable X tracks the sum of the numbers on the two upward faces.

The minimum possible sum is 2, which can only happen through a single combination: (1,1).

✅ Correct Answer Explanation

To find the probability of getting at most 6 heads (P(X≤6)) from 10 tosses of a fair coin, it is more efficient to calculate the complement probability of getting 7 or more heads and subtract it from 1: P(X≤6)=1-P(X≥7)=1-[P(X=7)+P(X=8)+P(X=9)+P(X=10)] The probability for any specific number of successes r with a fair coin (p=q=0.5) is given by P(X=r)=(10/r)/1024. Let's find the coefficients for the upper tail: (10/7)=120 (10/8)=45 (10/9)=10 (10/10)=1 Summing these tail outcomes gives: 120+45+10+1=176. Now, subtract this tail sum from the total number of possibilities (1024): P(X≤6)=1024-176/1024=848/1024 Simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 16: 848÷16/1024÷16=53/64 This matches Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 193/512 represents a different probability range and does not match the calculated value.
Option C → Incorrect because 105/512 is a smaller fraction that accounts for only a few core combinations.
Option D → Incorrect because while 848/1024 is numerically correct before simplifying, standard multiple-choice questions require choosing the fully simplified fraction option when available. Let's look closely at Option B vs D. 848/1024 is exactly equal to 53/64. Since both values are numerically identical, let's re-verify why B is highlighted. Option B provides the simplified form.

🎯 Strategy

Used: Direct Quantification

Application: Calculate the upper tail terms using the complement rule, then simplify the fraction.

Final Logic: The arithmetic leads directly to 848/1024, which reduces to 53/64 (Option B).

🧠 Memory Trick

Use the complement rule to save time. It's much faster to calculate the 4 terms from 7 to 10 and subtract from 1 than to add up the 7 terms from 0 to 6.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Cumulative Binomial Probabilities

20 Using the binomial defective basket model where n = 2 and p = 3/10, what is the probability of selecting AT LEAST ONE defective basket?

A 9/100

B 42/100

C 51/100

D 91/100

💡 Short Explanation

A random experiment is an action or process where the exact outcome cannot be known ahead of time.
Even though we know all the possible results in advance, the exact outcome of any single trial depends on chance.
Because you cannot predict which specific sweater will be drawn on a given turn, it fits the definition of a random experiment.

✅ Correct Answer Explanation

To find the probability of getting at least one defective basket (P(X≥1)) when selecting n=2 baskets, we use the complement rule. This rule states that the probability of an event happening at least once is equal to 1 minus the probability of it not happening at all: P(X≥1)=1-P(X=0) First, we find the probability of selecting a non-defective basket (q): q=1-p=1-3/10=7/10 Next, calculate the probability that both selected baskets are non-defective (X=0): P(X=0)=(2/0)⋅p^0⋅q^2=1⋅1⋅(7/10)^2=49/100 Finally, subtract this value from 1 to find the probability of getting at least one defective basket: P(X≥1)=1-49/100=100-49/100=51/100 This matches Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 9/100 is the probability of selecting exactly two defective baskets (p^2=(3/10)^2=9/100).
Option B → Incorrect because 42/100 is the probability of selecting exactly one defective basket ((2/1)pq=2×3/10×7/10=42/100).
Option D → Incorrect because 91/100 is a calculation error that does not match the complement of the non-defective squared fraction.

🎯 Strategy

Used: Direct Formula Application

Application: Apply the complement rule 1-q^2 to solve for "at least one" success in two trials.

Final Logic: Subtracting the all-failure probability (49/100) from 1 gives 51/100, confirming Option C.

🧠 Memory Trick

"At least one" is always 1-None. The chance of getting no defects is 7/10×7/10=49/100. Subtracting from 1 gives 51/100.

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Defective Item Sampling Models

CUET UG Applied Mathematics Booster Test 3 - Variance and Binomial Distribution

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