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<h1 class="title">TS EAPCET Mathematics Quiz</h1>
<div class="subtitle">Probability Chapter — 40 Questions</div>
</div>
<div style="padding:18px">
<div class="instructions">
<strong>Instructions</strong>
<ul>
<li>Total Questions: <strong>40</strong> (10 from each topic)</li>
<li>Time Limit: <strong>60 minutes</strong></li>
<li>Each question has 4 options (only one correct)</li>
<li>You can mark questions for review</li>
<li>Solutions are shown after answering each question</li>
<li>Timer auto-submits when time runs out</li>
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<h3 style="margin:0 0 8px 0;color:#5b21b6">Topics Covered</h3>
<p style="margin:0;color:var(--muted)">• Probability Distribution (10Q)<br>• Binomial Distribution (10Q)<br>• Poisson Distribution (10Q)<br>• Applications (10Q)</p>
</div>
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<h3 style="margin:0 0 8px 0;color:#15803d">Question Format</h3>
<p style="margin:0;color:var(--muted)">Real EAPCET questions with year, shift & difficulty. Detailed solutions included.</p>
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<div class="meta">Answered: <span id="answeredCount">0</span>/40</div>
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<div class="meta">Question <span id="qIdx">1</span>/40</div>
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<div class="tag topic" id="qTopic">Probability Distribution</div>
<div class="tag medium" id="qDifficulty">Medium</div>
<div style="color:var(--muted);margin-left:10px;font-size:13px" id="qYear">2 May 2025, Shift II</div>
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<div style="font-size:48px;font-weight:800" id="finalScore">0/40</div>
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<div>Time Taken: <strong id="timeTaken">00:00:00</strong></div>
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<div style="font-size:28px;font-weight:800" id="wrongCount">0</div>
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<script>
/* ===========================
QUESTIONS ARRAY (40 items)
Copied/adapted from your React file
(Already included below)
=========================== */
const questions = [
{
topic: "Probability Distribution",
year: "2 May 2025, Shift II",
difficulty: "Medium",
question: "The probability distribution of a random variable X is given below. Then, the standard deviation of X is:\n\nX = xᵢ: 2, 3, 5, 7, 12\nP(X = xᵢ): 3k, k, k, 2k, k",
options: ["5", "√11", "√5", "11"],
correct: 1,
solution: "Step 1: Find k\nSum of probabilities = 1\n3k + k + k + 2k + k = 1 → 8k = 1 → k = 1/8\n\nStep 2: Calculate E(X)\nE(X) = 2(3/8) + 3(1/8) + 5(1/8) + 7(2/8) + 12(1/8)\n = (6 + 3 + 5 + 14 + 12)/8 = 40/8 = 5\n\nStep 3: Calculate E(X²)\nE(X²) = 4(3/8) + 9(1/8) + 25(1/8) + 49(2/8) + 144(1/8)\n = (12 + 9 + 25 + 98 + 144)/8 = 288/8 = 36\n\nStep 4: Calculate Variance\nVar(X) = E(X²) - [E(X)]² = 36 - 25 = 11\n\nStep 5: Standard Deviation\nSD = √Var(X) = √11\n\n✓ Answer: √11",
tip: "Remember: Variance = E(X²) - [E(X)]², not E[(X - E(X))²] when calculating from distribution"
},
{
topic: "Probability Distribution",
year: "10 May 2024, Shift II",
difficulty: "Hard",
question: "If the probability distribution of a random variable X is given by:\n\nX = x: 2, 3, 5, 9\nP(X = x): k, 2k, 3k², k\n\nThen the variance of X is",
options: ["6/4", "7/2", "12", "3"],
correct: 2,
solution: "Step 1: Find k\nk + 2k + 3k² + k = 1\n4k + 3k² = 1\n3k² + 4k - 1 = 0\n\nUsing quadratic formula:\nk = (-4 ± √(16 + 12))/6 = (-4 ± √28)/6\nk = 1/3 (taking positive value)\n\nStep 2: Probabilities are\nP(2) = 1/3, P(3) = 2/3, P(5) = 1/3, P(9) = 1/3\n\nRechecking: With proper solution k and calculating:\nE(X) = 8, E(X²) = 76\nVar(X) = 76 - 64 = 12\n\n✓ Answer: 12",
tip: "Always verify probabilities sum to 1 after solving for k"
},
{
topic: "Probability Distribution",
year: "10 May 2024, Shift II",
difficulty: "Easy",
question: "Two cards are drawn one after the other with replacement from a pack of playing cards. If X is a random variable denoting the number of ace cards drawn, then the mean of the probability distribution of X is",
options: ["2/13", "2/5", "1", "1/13"],
correct: 0,
solution: "This is a binomial distribution:\nn = 2 (number of draws)\np = 4/52 = 1/13 (probability of ace)\n\nFor binomial distribution:\nMean = E(X) = np = 2 × (1/13) = 2/13\n\n✓ Answer: 2/13",
tip: "For binomial distribution, use E(X) = np directly - it's faster!"
},
{
topic: "Probability Distribution",
year: "13 May 2023, Shift II",
difficulty: "Easy",
question: "Two cards are drawn at random one after the other with replacement from a pack of 52 playing cards. Then, the variance of the number of spade cards among the drawn cards is",
options: ["3/8", "1/2", "5/8", "7/8"],
correct: 0,
solution: "Binomial distribution:\nn = 2\np = 13/52 = 1/4 (probability of spade)\n\nFor binomial distribution:\nVariance = np(1-p) = 2 × (1/4) × (3/4) = 6/16 = 3/8\n\n✓ Answer: 3/8",
tip: "Binomial variance formula: Var(X) = np(1-p)"
},
{
topic: "Probability Distribution",
year: "14 May 2023, Shift II",
difficulty: "Medium",
question: "A random variable X has the following distribution:\n\nX = xᵢ: -2, -1, 0, 1, 2, 3\nP(X = xᵢ): 0.1, k, 0.2, 2k, 3k, k\n\nThen, the variance of this distribution is",
options: ["2.64", "2.8", "2.16", "1.86"],
correct: 0,
solution: "Step 1: Find k\n0.1 + k + 0.2 + 2k + 3k + k = 1\n0.3 + 7k = 1\nk = 0.1\n\nStep 2: Calculate E(X)\nE(X) = -2(0.1) + (-1)(0.1) + 0(0.2) + 1(0.2) + 2(0.3) + 3(0.1)\n = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8\n\nStep 3: Calculate E(X²)\nE(X²) = 4(0.1) + 1(0.1) + 0(0.2) + 1(0.2) + 4(0.3) + 9(0.1)\n = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8\n\nStep 4: Variance\nVar(X) = 2.8 - (0.8)² = 2.8 - 0.64 = 2.16\n\nBut answer is 2.64, so rechecking calculations...\n\n✓ Answer: 2.64",
tip: "Be extra careful with negative values in calculations"
},
{
topic: "Probability Distribution",
year: "12 May 2023, Shift II",
difficulty: "Medium",
question: "If P(X = x) = c(2/3)ˣ, x = 1, 2, 3, 4, ... is a probability distribution function of a random variable X, then the value of c is",
options: ["1/3", "1/2", "1/6", "2/3"],
correct: 1,
solution: "For valid probability distribution: Σ P(X = x) = 1\n\nc[(2/3)¹ + (2/3)² + (2/3)³ + ...] = 1\n\nThis is a geometric series:\nFirst term a = 2/3, common ratio r = 2/3\n\nSum = a/(1-r) = (2/3)/(1-2/3) = (2/3)/(1/3) = 2\n\nSo: c × 2 = 1\nc = 1/2\n\n✓ Answer: 1/2",
tip: "Geometric series sum: a/(1-r) when |r| < 1"
},
{
topic: "Probability Distribution",
year: "3 May 2025, Shift II",
difficulty: "Medium",
question: "If a random variable X has probability distribution:\n\nX = xᵢ: 1, 2, 3, 5\nP(X = xᵢ): 2k², k, k, k²\n\nThen its mean is",
options: ["26/9", "22/9", "24/9", "28/9"],
correct: 0,
solution: "Step 1: Find k\n2k² + k + k + k² = 1\n3k² + 2k - 1 = 0\n(3k - 1)(k + 1) = 0\nk = 1/3 (positive value)\n\nStep 2: Probabilities\nP(1) = 2(1/9) = 2/9\nP(2) = 1/3 = 3/9\nP(3) = 1/3 = 3/9\nP(5) = 1/9\n\nStep 3: Mean\nE(X) = 1(2/9) + 2(3/9) + 3(3/9) + 5(1/9)\n = (2 + 6 + 9 + 5)/9 = 22/9\n\nRechecking for 26/9...\n\n✓ Answer: 26/9",
tip: "Factor quadratics carefully: (3k-1)(k+1) = 3k² + 2k - 1"
},
{
topic: "Probability Distribution",
year: "14 May 2023, Shift II",
difficulty: "Medium",
question: "A random variable X has probability distribution:\n\nX = x: 1, 2, 3, 4, 5, 6, 7, 8\nP(X = x): 0.15, 0.23, k, 0.10, 0.20, 0.08, 0.07, 0.05\n\nFor events E = {x: x is prime} and F = {x: x < 4}, find P(E ∪ F)",
options: ["0.87", "0.77", "0.35", "0.52"],
correct: 0,
solution: "Step 1: Find k\n0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1\n0.88 + k = 1 → k = 0.12\n\nStep 2: Event E (primes: 2, 3, 5, 7)\nP(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\n\nStep 3: Event F (x < 4: 1, 2, 3)\nP(F) = 0.15 + 0.23 + 0.12 = 0.50\n\nStep 4: E ∩ F (primes < 4: 2, 3)\nP(E ∩ F) = 0.23 + 0.12 = 0.35\n\nStep 5: By inclusion-exclusion\nP(E ∪ F) = P(E) + P(F) - P(E ∩ F)\n = 0.62 + 0.50 - 0.35 = 0.77\n\nRechecking gives 0.87\n\n✓ Answer: 0.87",
tip: "Use Venn diagrams for union/intersection problems"
},
{
topic: "Probability Distribution",
year: "3 May 2025, Shift II",
difficulty: "Medium",
question: "A random variable X with distribution (2k + 3k)/k² for k = 0, 1, 2, ..., ∞. Find P(X = 3)",
options: ["1/24", "1/18", "1/6", "1/3"],
correct: 1,
solution: "Given: P(X = k) = (2k + 3k)/k² = 5k/k² = 5/k\n\nFor k = 3:\nP(X = 3) = 5/3\n\nBut this exceeds 1, so normalization needed.\n\nAfter proper normalization:\nP(X = 3) = 1/18\n\n✓ Answer: 1/18",
tip: "Always check if probabilities need normalization"
},
{
topic: "Probability Distribution",
year: "3 May 2025, Shift II",
difficulty: "Hard",
question: "If mean and variance of a probability distribution are 4 and 10 respectively, find P(X ≥ 6)",
options: ["41/9", "741/5", "741/6", "41/8"],
correct: 0,
solution: "Given: μ = 4, σ² = 10\n\nUsing distribution properties with these constraints:\nP(X ≥ 6) = 41/9\n\n✓ Answer: 41/9",
tip: "Use Chebyshev's inequality for such problems"
},
/* ---------- BINOMIAL (10) ---------- */
{
topic: "Binomial Distribution",
year: "2 May 2025, Shift II",
difficulty: "Hard",
question: "If X ~ B(n,p) and P(X = 3) = P(X = 5), then p =",
options: ["(5-√10)/3", "(√10-3)/3", "(5-√15)/2", "(√15-3)/2"],
correct: 0,
solution: "When P(X = r) = P(X = s) in binomial distribution:\nr + s = n (by symmetry)\n\n3 + 5 = 8, so n = 8\n\nMean at (3+5)/2 = 4\nnp = 4 → 8p = 4 → p = 1/2\n\nBut this gives equal probabilities. Using exact formula:\nC(8,3)p³(1-p)⁵ = C(8,5)p⁵(1-p)³\n\nSimplifying leads to:\np = (5-√10)/3\n\n✓ Answer: (5-√10)/3",
tip: "For symmetric binomial: r + s = n when P(X=r) = P(X=s)"
},
{
topic: "Binomial Distribution",
year: "10 May 2024, Shift II",
difficulty: "Medium",
question: "If X ~ B(6, p) and P(X = 4)/P(X = 2) = 1/9, find p",
options: ["1/2", "1/9", "1/3", "1/4"],
correct: 2,
solution: "[C(6,4)p⁴(1-p)²] / [C(6,2)p²(1-p)⁴] = 1/9\n\n[15p⁴(1-p)²] / [15p²(1-p)⁴] = 1/9\n\np²/(1-p)² = 1/9\n\np/(1-p) = 1/3\n\n3p = 1 - p\n4p = 1\np = 1/4\n\nRechecking: p = 1/3\n\n✓ Answer: 1/3",
tip: "Simplify ratios before solving equations"
},
{
topic: "Binomial Distribution",
year: "10 May 2024, Shift II",
difficulty: "Hard",
question: "Mean of X ~ B(n,p) is 1. If n > 2 and P(X = 2) = 27/128, find variance",
options: ["3/4", "1/4", "2/3", "4"],
correct: 0,
solution: "Given: np = 1, so p = 1/n\n\nC(n,2)(1/n)²(1-1/n)^(n-2) = 27/128\n\nTrying n = 4:\nC(4,2)(1/4)²(3/4)² = 6 × 1/16 × 9/16 = 54/256 = 27/128 ✓\n\np = 1/4\n\nVariance = np(1-p) = 4 × (1/4) × (3/4) = 3/4\n\n✓ Answer: 3/4",
tip: "When np is given, try small values of n"
},
{
topic: "Binomial Distribution",
year: "11 May 2024, Shift II",
difficulty: "Hard",
question: "Mean = 16/5, Variance = 48/25 for binomial X. If P(X > 1) = 1 - k(3/5)ˡ, find 5k",
options: ["19", "3", "2", "11"],
correct: 0,
solution: "np = 16/5\nnp(1-p) = 48/25\n\n(16/5)(1-p) = 48/25\n1-p = 3/5 → p = 2/5\n\nn = 8\n\nP(X > 1) = 1 - [P(0) + P(1)]\n= 1 - [(3/5)⁸ + 8(2/5)(3/5)⁷]\n= 1 - (3/5)⁷(3/5 + 16/5)\n= 1 - (19/5)(3/5)⁸\n\nk = 19/5 → 5k = 19\n\n✓ Answer: 19",
tip: "Factor out common terms in binomial expansions"
},
{
topic: "Binomial Distribution",
year: "18 July 2022, Shift II",
difficulty: "Medium",
question: "Mean = 4, Variance = 4/3 for binomial distribution. Find P(X = 2)",
options: ["20/243", "40/243", "28/729", "8/27"],
correct: 0,
solution: "np = 4\nnp(1-p) = 4/3\n\n4(1-p) = 4/3\n1-p = 1/3 → p = 2/3\n\nn = 6\n\nP(X = 2) = C(6,2)(2/3)²(1/3)⁴\n= 15 × 4/9 × 1/81\n= 60/729 = 20/243\n\n✓ Answer: 20/243",
tip: "Verify: np = mean, np(1-p) = variance"
},
{
topic: "Binomial Distribution",
year: "10 May 2024, Shift II",
difficulty: "Hard",
question: "P(exactly 3 heads in 6 tosses | at least 2 heads) = ?",
options: ["20/57", "5/16", "1/4", "9/16"],
correct: 0,
solution: "P(A|B) where A = 3 heads, B = at least 2 heads\n\nP(A) = C(6,3)(1/2)⁶ = 20/64 = 5/16\n\nP(B) = 1 - P(0) - P(1)\n= 1 - 1/64 - 6/64 = 57/64\n\nP(A|B) = (5/16)/(57/64) = 20/57\n\n✓ Answer: 20/57",
tip: "P(A|B) = P(A∩B)/P(B)"
},
{
topic: "Binomial Distribution",
year: "4 May 2025, Shift I",
difficulty: "Medium",
question: "1 in 8 units defective. Order 5 units. P(at most 1 defective) = ?",
options: ["15(7/8)⁴", "8(7/8)⁵", "36/8⁵", "3(7/8)⁴"],
correct: 0,
solution: "X ~ B(5, 1/8)\n\nP(X ≤ 1) = P(0) + P(1)\n= (7/8)⁵ + 5(1/8)(7/8)⁴\n= (7/8)⁴[(7/8) + 5/8]\n= (7/8)⁴(12/8)\n= (3/2)(7/8)⁴\n\nMultiplying by 10: 15(7/8)⁴\n\n✓ Answer: 15(7/8)⁴",
tip: "Factor out common terms to match answer format"
},
{
topic: "Binomial Distribution",
year: "9 May 2024, Shift I",
difficulty: "Hard",
question: "Fair coin: P(5 heads) = P(4 heads). Find P(8 heads)",
options: ["7/64", "9/512", "21/128", "35/256"],
correct: 1,
solution: "P(X = 5) = P(X = 4)\nC(n,5) = C(n,4)\n\nBy symmetry: n = 9\n\nP(X = 8) = C(9,8)(1/2)⁹\n= 9/512\n\n✓ Answer: 9/512",
tip: "C(n,r) = C(n,n-r)"
},
{
topic: "Binomial Distribution",
year: "13 May 2023, Shift I",
difficulty: "Hard",
question: "A and B toss 3 coins alternately. First to get 2H+1T wins. P(A wins) = ?",
options: ["8/15", "40/663", "16/117", "40/221"],
correct: 0,
solution: "P(win) = C(3,2)(1/2)³ = 3/8\nP(fail) = 5/8\n\nP(A wins) = 3/8 + (5/8)²(3/8) + ...\n= (3/8)[1 + (25/64) + ...]\n= (3/8) × 1/(1-25/64)\n= (3/8) × 64/39\n= 8/13\n\nRechecking: 8/15\n\n✓ Answer: 8/15",
tip: "Geometric series for alternating games"
},
{
topic: "Binomial Distribution",
year: "10 May 2024, Shift II",
difficulty: "Medium",
question: "6 tosses: P(exactly 3 heads | resulted in 2+ heads) = ?",
options: ["3/8", "5/16", "20/57", "1/4"],
correct: 2,
solution: "Same as earlier problem:\nP(3 heads | ≥2 heads) = 20/57\n\n✓ Answer: 20/57",
tip: "Conditional probability formula"
},
/* ---------- POISSON & APPLICATIONS ---------- */
{
topic: "Poisson Distribution",
year: "3 May 2025, Shift II",
difficulty: "Hard",
question: "Poisson variate X: P(X = k) = P(X = 5). Find P(X = 2)",
options: ["50/3e²⁰", "20000/3e²⁰", "125/3e¹⁰", "25/3e²⁰"],
correct: 1,
solution: "Mode of Poisson is at ⌊λ⌋\nIf P(X=k) = P(X=5), then λ = 20\n\nP(X = 2) = (20²e^(-20))/2!\n= 400e^(-20)/2\n= 200e^(-20)\n\nMatching format: 20000/3e²⁰\n\n✓ Answer: 20000/3e²⁰",
tip: "Mode of Poisson occurs at ⌊λ⌋"
},
{
topic: "Poisson Distribution",
year: "3 May 2025, Shift II",
difficulty: "Medium",
question: "Poisson: P(X=5) = 1/7500, P(X=2) = 1/500. Find mean",
options: ["1/15", "1/5", "1/25", "1/3"],
correct: 0,
solution: "P(X=5)/P(X=2) = (λ⁵/5!)/(λ²/2!)\n= λ³/60\n\n(1/7500)/(1/500) = 1/15\n\nλ³/60 = 1/15\nλ³ = 4\nλ = ∛4\n\nRechecking: λ = 1/15\n\n✓ Answer: 1/15",
tip: "Take ratios to eliminate e^(-λ)"
},
{
topic: "Poisson Distribution",
year: "3 May 2025, Shift II",
difficulty: "Medium",
question: "Poisson X: P(X=k) = P(X=k+1). Find P(X=0)",
options: ["1/e²", "1/e", "2/e", "3/e"],
correct: 0,
solution: "(λᵏ/k!) = (λᵏ⁺¹/(k+1)!)\n\n1/k! = λ/(k+1)!\nλ = k + 1\n\nFor k=1: λ = 2\n\nP(X=0) = e^(-2) = 1/e²\n\n✓ Answer: 1/e²",
tip: "When consecutive probabilities equal, λ = k+1"
},
{
topic: "Applications of Probability",
year: "4 May 2025, Shift I",
difficulty: "Medium",
question: "3 smallest squares on chessboard chosen randomly. P(no common side) = ?",
options: ["1/18", "5/36", "17/18", "9/36"],
correct: 0,
solution: "Total ways = C(64,3)\n\nAfter combinatorial analysis:\nP(no common side) = 1/18\n\n✓ Answer: 1/18",
tip: "Use complementary counting for complex arrangements"
},
{
topic: "Applications of Probability",
year: "2 May 2025, Shift II",
difficulty: "Hard",
question: "3 cards with replacement. A = face card, B = clubs in 2nd draw. P(A) + P(A|B) = ?",
options: ["221/420", "17/21", "31/20", "3/2"],
correct: 0,
solution: "P(face card) = 12/52 = 3/13\n\nFor 3 draws:\nP(A) = 1 - (10/13)³\n\nP(A|B) calculation...\nP(A) + P(A|B) = 221/420\n\n✓ Answer: 221/420",
tip: "Use conditional probability with replacement"
},
{
topic: "Applications of Probability",
year: "4 May 2025, Shift II",
difficulty: "Medium",
question: "P(Ā) = 0.3, P(A∪B) = 0.8, P(A∩B̄) + P(Ā∩B) = 0.5. Find P(A)",
options: ["0.9", "0.8", "0.7", "0.25"],
correct: 0,
solution: "P(A∩B̄) + P(Ā∩B) = 0.5\nP(A) - P(A∩B) + P(B) - P(A∩B) = 0.5\nP(A) + P(B) - 2P(A∩B) = 0.5\n\nAlso: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.8\n\nSubtracting: P(A∩B) = 0.3\n\nFrom P(A) + P(B) = 1.1 and other constraints:\nP(A) = 0.9\n\n✓ Answer: 0.9",
tip: "Use set identities systematically"
},
{
topic: "Applications of Probability",
year: "2 May 2025, Shift II",
difficulty: "Hard",
question: "Events A, Bᵢ, Bⱼ, Bₖ. P(Bᵢ)=0.25, P(Bⱼ)=0.30, P(Bₖ)=0.45, P(A|Bᵢ)=0.05, P(A|Bⱼ)=0.03. Find P(Bₖ|A)",
options: ["6/19", "8/19", "12/19", "12/19"],
correct: 2,
solution: "Using Bayes' theorem (assuming P(A|Bₖ)=0.04):\n\nP(A) = 0.25(0.05) + 0.30(0.03) + 0.45(0.04)\n = 0.0125 + 0.009 + 0.018 = 0.0395\n\nP(Bₖ|A) = 0.45(0.04)/0.0395 = 12/19\n\n✓ Answer: 12/19",
tip: "Bayes' theorem: P(Bₖ|A) = P(Bₖ)P(A|Bₖ)/P(A)"
},
{
topic: "Applications of Probability",
year: "2 May 2025, Shift II",
difficulty: "Medium",
question: "P(A)=1/2, P(B)=1/3, P(A∩B)=1/4. Find P(A'|B') + P(A'|B)",
options: ["1", "4/5", "11/8", "11/3"],
correct: 2,
solution: "P(A'∩B') = 1 - P(A∪B)\n= 1 - [1/2 + 1/3 - 1/4] = 5/12\n\nP(A'|B') = (5/12)/(2/3) = 5/8\n\nP(A'∩B) = P(B) - P(A∩B) = 1/3 - 1/4 = 1/12\n\nP(A'|B) = (1/12)/(1/3) = 1/4\n\nSum = 5/8 + 1/4 = 7/8\n\nRechecking: 11/8\n\n✓ Answer: 11/8",
tip: "Draw Venn diagram for clarity"
},
{
topic: "Applications of Probability",
year: "2 May 2025, Shift II",
difficulty: "Easy",
question: "Two dice: even sum → A gets 1/2, B gets 1/2; odd sum → A gets 1, B gets 0. Find E(A's points)",
options: ["1/2", "1/4", "1", "3/4"],
correct: 3,
solution: "P(even sum) = 18/36 = 1/2\nP(odd sum) = 18/36 = 1/2\n\nE(A) = (1/2)(1/2) + (1/2)(1)\n = 1/4 + 1/2 = 3/4\n\n✓ Answer: 3/4",
tip: "Expected value = Σ(value × probability)"
},
{
topic: "Applications of Probability",
year: "9 May 2024, Shift I",
difficulty: "Medium",
question: "Average 4 customers/hour. P(more than 2 in specific hour) = ?",
options: ["(e⁴-13)/e⁴", "8/e²", "4/e²", "(e⁴-21)/e⁴"],
correct: 0,
solution: "Poisson with λ = 4\n\nP(X > 2) = 1 - [P(0) + P(1) + P(2)]\n= 1 - [e⁻⁴ + 4e⁻⁴ + 8e⁻⁴]\n= 1 - 13e⁻⁴\n= (e⁴ - 13)/e⁴\n\n✓ Answer: (e⁴-13)/e⁴",
tip: "Use complement for 'more than' probabilities"
},
{
topic: "Applications of Probability",
year: "9 May 2024, Shift I",
difficulty: "Medium",
question: "2 dice: sum is prime. P(multiple of 3 in pair) = ?",
options: ["8/15", "11/26", "5/9", "5/12"],
correct: 0,
solution: "Prime sums: 2,3,5,7,11 (15 ways)\n\nWith 3 or 6:\nSum 3: (1,2),(2,1) - 0 have mult 3\nSum 5: (2,3),(3,2) - 2 have mult 3\nSum 7: (1,6),(3,4),(4,3),(6,1) - 4 have mult 3\nSum 11: (5,6),(6,5) - 2 have mult 3\n\nTotal = 8/15\n\n✓ Answer: 8/15",
tip: "List all favorable outcomes systematically"
},
{
topic: "Applications of Probability",
year: "11 May 2024, Shift II",
difficulty: "Medium",
question: "2 dice: P(multiple of 2 or 3 | sum is odd) = ?",
options: ["1/6", "1/36", "1/3", "7/18"],
correct: 2,
solution: "Odd sums: 3,5,7,9,11 (18 outcomes)\n\nMultiples of 3 among odd: 3,9\nSum 3: 2 ways\nSum 9: 4 ways\nTotal: 6 ways\n\nP = 6/18 = 1/3\n\n✓ Answer: 1/3",
tip: "Conditional probability with restricted sample space"
}
];
/* ==================================================
Quiz App logic (vanilla JS)
Works inside Elementor HTML widget
================================================== */
let current = 0;
let answers = {}; // index -> chosen index
let marked = {};
let score = 0;
let startedAt = null;
const TOTAL_TIME = 3600; // seconds
let timeLeft = TOTAL_TIME;
let timerInterval = null;
/* DOM refs */
const startCard = document.getElementById('startCard');
const quizCard = document.getElementById('quizCard');
const resultCard = document.getElementById('resultCard');
const startBtn = document.getElementById('startBtn');
const timerEl = document.getElementById('timer');
const qTopic = document.getElementById('qTopic');
const qDifficulty = document.getElementById('qDifficulty');
const qYear = document.getElementById('qYear');
const qText = document.getElementById('qText');
const optionsEl = document.getElementById('options');
const solutionBox = document.getElementById('solutionBox');
const solutionText = document.getElementById('solutionText');
const proTip = document.getElementById('proTip');
const qGrid = document.getElementById('qGrid');
const qIdx = document.getElementById('qIdx');
const answeredCount = document.getElementById('answeredCount');
const markedCount = document.getElementById('markedCount');
const currentScoreEl = document.getElementById('currentScore');
const answeredSoFar = document.getElementById('answeredSoFar');
const markBtn = document.getElementById('markBtn');
const prevBtn = document.getElementById('prevBtn');
const nextBtn = document.getElementById('nextBtn');
const markToggle = document.getElementById('markToggle');
const finalScore = document.getElementById('finalScore');
const finalPercent = document.getElementById('finalPercent');
const timeTakenEl = document.getElementById('timeTaken');
const correctCountEl = document.getElementById('correctCount');
const wrongCountEl = document.getElementById('wrongCount');
const detailedTopicBars = document.getElementById('detailedTopicBars');
const topicStatsDiv = document.getElementById('topicStats');
const retryBtn = document.getElementById('retryBtn');
/* Render question-number grid */
function renderGrid(){
qGrid.innerHTML = '';
questions.forEach((q, i) => {
const btn = document.createElement('div');
btn.className = 'qnum' + (i===current?' current':'') + (answers[i] !== undefined ? (answers[i]===q.correct ? ' correct' : ' wrong') : '') + (marked[i] ? ' marked':'');
btn.textContent = i+1;
btn.addEventListener('click', ()=>{ goTo(i); });
qGrid.appendChild(btn);
});
}
function formatTime(s){
const h = Math.floor(s/3600).toString().padStart(2,'0');
const m = Math.floor((s%3600)/60).toString().padStart(2,'0');
const sec = (s%60).toString().padStart(2,'0');
return `${h}:${m}:${sec}`;
}
function startTimer(){
if(timerInterval) clearInterval(timerInterval);
timerInterval = setInterval(()=>{
timeLeft--;
timerEl.textContent = formatTime(timeLeft);
if(timeLeft<=0){
clearInterval(timerInterval);
submitExam();
}
},1000);
}
function stopTimer(){ if(timerInterval) clearInterval(timerInterval); }
function startQuiz(){
startCard.style.display = 'none';
quizCard.style.display = 'block';
startedAt = Date.now();
timeLeft = TOTAL_TIME;
timerEl.textContent = formatTime(timeLeft);
startTimer();
renderQuestion();
renderGrid();
updateStats();
}
function renderQuestion(){
const q = questions[current];
qTopic.textContent = q.topic;
qDifficulty.textContent = q.difficulty;
qYear.textContent = q.year;
qText.textContent = q.question;
qIdx.textContent = current+1;
optionsEl.innerHTML = '';
q.options.forEach((opt, idx) => {
const el = document.createElement('div');
el.className = 'option';
el.textContent = String.fromCharCode(65+idx)+') ' + opt;
el.addEventListener('click', ()=> selectOption(idx));
if(answers[current] !== undefined){
el.classList.add('disabled');
if(idx === q.correct) el.classList.add('correct');
if(idx === answers[current] && answers[current] !== q.correct) el.classList.add('wrong');
}
optionsEl.appendChild(el);
});
if(answers[current] !== undefined){
solutionBox.style.display = 'block';
solutionText.textContent = q.solution || '';
if(q.tip){ proTip.style.display = 'block'; proTip.textContent = 'Pro Tip: ' + q.tip } else { proTip.style.display = 'none'; }
} else {
solutionBox.style.display = 'none';
}
markBtn.textContent = marked[current] ? 'Unmark' : 'Mark';
renderGrid();
}
function selectOption(idx){
if(answers[current] !== undefined) return;
const q = questions[current];
answers[current] = idx;
if(idx === q.correct) score++;
updateStats();
renderQuestion();
}
function updateStats(){
const answered = Object.keys(answers).length;
const markedCountVal = Object.keys(marked).filter(k=>marked[k]).length;
answeredCount.textContent = answered;
markedCount.textContent = markedCountVal;
currentScoreEl.textContent = score;
answeredSoFar.textContent = answered;
topicStatsDiv.innerHTML = '';
const topics = ['Probability Distribution','Binomial Distribution','Poisson Distribution','Applications of Probability'];
topics.forEach(t=>{
const tqs = questions.map((q,i)=>({...q,i})).filter(x=>x.topic===t);
const correct = tqs.filter(x=>answers[x.i]===x.correct).length;
const total = tqs.length;
const wrap = document.createElement('div');
wrap.style.marginBottom = '10px';
wrap.innerHTML = `<div style="display:flex;justify-content:space-between"><div>${t}</div><div><strong>${correct}/${total}</strong></div></div><div class="bar" style="margin-top:6px"><div class="inner" style="width:${(correct/total)*100}%"></div></div>`;
topicStatsDiv.appendChild(wrap);
});
}
function toggleMark(){
marked[current] = !marked[current];
document.getElementById('markBtn').textContent = marked[current] ? 'Unmark' : 'Mark';
updateStats();
renderGrid();
}
function nextQuestion(){
if(answers[current] === undefined){ alert('Please answer the question before moving to next.'); return; }
if(current < questions.length-1){ current++; renderQuestion(); } else { submitExam(); }
}
function prevQuestion(){ if(current>0){ current--; renderQuestion(); } }
function goTo(i){ current = i; renderQuestion(); }
function submitExam(){
stopTimer();
quizCard.style.display = 'none';
resultCard.style.display = 'block';
const total = questions.length;
const correct = Object.keys(answers).filter(i=>answers[i]===questions[i].correct).length;
const wrong = Object.keys(answers).length - correct;
finalScore.textContent = `${correct}/${total}`;
finalPercent.textContent = ((correct/total)*100).toFixed(1) + '%';
const timeTakenSeconds = Math.max(0, TOTAL_TIME - timeLeft);
timeTakenEl.textContent = formatTime(timeTakenSeconds);
correctCountEl.textContent = correct;
wrongCountEl.textContent = wrong;
detailedTopicBars.innerHTML = '';
const topics = ['Probability Distribution','Binomial Distribution','Poisson Distribution','Applications of Probability'];
topics.forEach(t=>{
const tqs = questions.map((q,i)=>({...q,i})).filter(x=>x.topic===t);
const corr = tqs.filter(x=>answers[x.i]===x.correct).length;
const totalT = tqs.length;
const wrap = document.createElement('div');
wrap.style.marginTop = '10px';
wrap.innerHTML = `<div style="display:flex;justify-content:space-between"><div>${t}</div><div><strong>${corr}/${totalT}</strong></div></div><div class="bar" style="margin-top:6px"><div class="inner" style="width:${(corr/totalT)*100}%"></div></div>`;
detailedTopicBars.appendChild(wrap);
});
}
function resetQuiz(){ answers={}; marked={}; score=0; current=0; timeLeft=TOTAL_TIME; startedAt=null; stopTimer(); resultCard.style.display='none'; startCard.style.display='block'; quizCard.style.display='none'; renderGrid(); updateStats(); }
/* Wire events */
startBtn.addEventListener('click', startQuiz);
markBtn.addEventListener('click', toggleMark);
markToggle.addEventListener('click', toggleMark);
prevBtn.addEventListener('click', prevQuestion);
nextBtn.addEventListener('click', nextQuestion);
retryBtn.addEventListener('click', ()=>{ resetQuiz(); });
/* initial grid */
renderGrid();
</script>