### Related Formula Probability of non-zero outcome: P(E) = 1 - P(E^prime)$P(E) = 1 - P(E^\prime)$ where E^prime$E^\prime$ represents the condition i^k_1 + i^k_2 = 0$i^{k_1} + i^{k_2} = 0$. ### Core Logic Powers of i$i$ repeat periodically every 4 cycles: \i, -1, -i, 1\$\{i, -1, -i, 1\}$. Total possible outcomes for the pair (i^k_1, i^k_2)$(i^{k_1}, i^{k_2})$ are 4 times 4 = 16$4 \times 4 = 16$ cases. ### Step 1: Finding Unfavorable Cases The sum is zero (i^k_1 + i^k_2 = 0$i^{k_1} + i^{k_2} = 0$) when the values are additive inverses: 1. (1, -1)$(1, -1)$ 2. (-1, 1)$(-1, 1)$ 3. (i, -i)$(i, -i)$ 4. (-i, i)$(-i, i)$ This yields exactly 4 unfavorable cases. ### Step 2: Computing Probability textProbability = frac16 - 416 = frac1216 = frac34$\text{Probability} = \frac{16 - 4}{16} = \frac{12}{16} = \frac{3}{4}$ ### Pattern Recognition Whenever modular cycles exist (like exponents of i$i$ mod 4), compress infinite natural choices safely into a single cycle map. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers Class 12 Maths: Probability

### Related Formula Variance formula for discrete probability distributions: sigma^2 = sum p_i x_i^2 - mu^2$\sigma^2 = \sum p_i x_i^2 - \mu^2$ ### Core Logic Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good):

### Step 1: Calculating the Mean mu = sum x_i p_i = 0left(frac4290 ight) + 1left(frac4290 ight) + 2left(frac690 ight) = frac5490 = frac35$\mu = \sum x_i p_i = 0\left(\frac{42}{90} ight) + 1\left(\frac{42}{90} ight) + 2\left(\frac{6}{90} ight) = \frac{54}{90} = \frac{3}{5}$ ### Step 2: Evaluating the Variance Metric sigma^2 = sum p_i x_i^2 - mu^2 = left[0 + 1^2left(frac4290 ight) + 2^2left(frac690 ight)right] - left(frac35 ight)^2$\sigma^2 = \sum p_i x_i^2 - \mu^2 = \left[0 + 1^2\left(\frac{42}{90} ight) + 2^2\left(\frac{6}{90} ight)\right] - \left(\frac{3}{5} ight)^2$ sigma^2 = frac6690 - frac925 = frac1115 - frac925 = frac2875$\sigma^2 = \frac{66}{90} - \frac{9}{25} = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$ ### Pattern Recognition Discrete tables are best managed by computing component factor rows systematically before finalizing variance updates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Related Formula Sum of geometric sequence series configuration: S_N = fraca(r^N - 1)r - 1$S_N = \frac{a(r^N - 1)}{r - 1}$ Total counts of permutations of 5 unique letter combinations = 5! = 120$= 5! = 120$. ### Core Logic Set initial base state configuration parameter P(W_1) = x$P(W_1) = x$ [cite: 1431]. Since probabilities escalate geometrically [cite: 1433]: sum_i=1^120 P(W_i) = x + 2x + 2^2x + dots + 2^119x = 1$\sum_{i=1}^{120} P(W_i) = x + 2x + 2^2x + \dots + 2^{119}x = 1$ [cite: 1432, 1433] x cdot frac2^120 - 12 - 1 = 1 implies x = frac12^120 - 1$x \cdot \frac{2^{120} - 1}{2 - 1} = 1 \implies x = \frac{1}{2^{120} - 1}$ [cite: 1434] ### Step 1: Finding Dictionary Rank of CDBEA Calculate alphabetical sequential position index values[cite: 1437]: - Words starting with A$A$: 4! = 24$4! = 24$ [cite: 1438] - Words starting with B$B$: 4! = 24$4! = 24$ [cite: 1438] - Words starting with CA$CA$: 3! = 6$3! = 6$ [cite: 1438] - Words starting with CB$CB$: 3! = 6$3! = 6$ [cite: 1439] - Words starting with CDA$CDA$: 2! = 2$2! = 2$ [cite: 1439] - Next word alphabetically: CDBAE$CDBAE$ (Rank 63) [cite: 1439] - Target word: CDBEA$CDBEA$ (Rank 64) [cite: 1440] Thus, target word index is exactly n = 64$n = 64$[cite: 1442]. ### Step 2: Probabilistic coefficient calculation Evaluate the specific targeted weighted entry value [cite: 1442]: P(W_64) = 2^63 cdot P(W_1) = frac2^632^120-1$P(W_{64}) = 2^{63} \cdot P(W_1) = \frac{2^{63}}{2^{120}-1}$ [cite: 1442, 1443] Matching structural values with expression flags [cite: 1443]: alpha = 63, quad beta = 120$\alpha = 63, \quad \beta = 120$ [cite: 1443] alpha + beta = 63 + 120 = 183$\alpha + \beta = 63 + 120 = 183$ [cite: 1443] ### Pattern Recognition When probability states map through clean binary power scale jumps, their total summation creates standard Mersenne number forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability

### Core Logic Let E_1$E_1$ be the event that the lost card is a spade, and E_2$E_2$ be the event that the lost card is not a spade. P(E_1) = frac1352 = frac14 quad textand quad P(E_2) = frac3952 = frac34$P(E_1) = \frac{13}{52} = \frac{1}{4} \quad \text{and} \quad P(E_2) = \frac{39}{52} = \frac{3}{4}$ Let A$A$ be the event that n$n$ cards drawn from the remaining 51$51$ cards are all spades. - If the lost card was a spade (E_1$E_1$), there are 12$12$ spades left out of 51$51$: P(A|E_1) = fracbinom12nbinom51n$P(A|E_1) = \frac{\binom{12}{n}}{\binom{51}{n}}$ - If the lost card was not a spade (E_2$E_2$), there are 13$13$ spades left out of 51$51$: P(A|E_2) = fracbinom13nbinom51n$P(A|E_2) = \frac{\binom{13}{n}}{\binom{51}{n}}$ ### Step 1: Applying Bayes Theorem We are given the posterior probability that the lost card is a spade, P(E_1|A) = frac1150$P(E_1|A) = \frac{11}{50}$: P(E_1|A) = fracP(E_1)P(A|E_1)P(E_1)P(A|E_1) + P(E_2)P(A|E_2) = frac1150$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{11}{50}$ fracfrac14 cdot fracbinom12nbinom51nfrac14 cdot fracbinom12nbinom51n + frac34 cdot fracbinom13nbinom51n = frac1150$\frac{\frac{1}{4} \cdot \frac{\binom{12}{n}}{\binom{51}{n}}}{\frac{1}{4} \cdot \frac{\binom{12}{n}}{\binom{51}{n}} + \frac{3}{4} \cdot \frac{\binom{13}{n}}{\binom{51}{n}}} = \frac{11}{50}$ Canceling out the shared fractions frac14$\frac{1}{4}$ and binom51n$\binom{51}{n}$: fracbinom12nbinom12n + 3binom13n = frac1150$\frac{\binom{12}{n}}{\binom{12}{n} + 3\binom{13}{n}} = \frac{11}{50}$ ### Step 2: Simplifying Binomial Coefficients Express binom13n$\binom{13}{n}$ in terms of binom12n$\binom{12}{n}$ using the identity binom13n = frac1313-nbinom12n$\binom{13}{n} = \frac{13}{13-n}\binom{12}{n}$: fracbinom12nbinom12n + 3 cdot left[ frac1313-n binom12n right] = frac1150$\frac{\binom{12}{n}}{\binom{12}{n} + 3 \cdot \left[ \frac{13}{13-n} \binom{12}{n} \right]} = \frac{11}{50}$ Canceling binom12n$\binom{12}{n}$ from numerator and denominator: frac11 + frac3913-n = frac1150 implies frac13-n13-n + 39 = frac1150$\frac{1}{1 + \frac{39}{13-n}} = \frac{11}{50} \implies \frac{13-n}{13-n + 39} = \frac{11}{50}$ frac13-n52-n = frac1150$\frac{13-n}{52-n} = \frac{11}{50}$ Cross-multiplying: 50(13 - n) = 11(52 - n)$50(13 - n) = 11(52 - n)$ 650 - 50n = 572 - 11n implies 39n = 78 implies n = 2$650 - 50n = 572 - 11n \implies 39n = 78 \implies n = 2$ ### Pattern Recognition When expanding binomial dynamic ratios like fracbinomAnbinomA+1n$\frac{\binom{A}{n}}{\binom{A+1}{n}}$, always reduce the larger term fractionally using the absorption property to cancel out the factorial variables quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula Classical Probability: textP(E) = fractextFavorable CasestextTotal Cases = fracn(E)n(S)$\text{P}(E) = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{n(E)}{n(S)}$ ### Core Logic Total pool size = 4 + 2 + 10 = 16$= 4 + 2 + 10 = 16$ people. We choose 12. Total cases n(S) = ^16mathrmC_12 = 1820$n(S) = {}^{16}\mathrm{C}_{12} = 1820$. List distinct combinations satisfying 'at least 3 engineers' and 'at least 1 doctor': 1) 3 Engineers, 1 Doctor, 8 Professors: ^4mathrmC_3 times ^2mathrmC_1 times ^10mathrmC_8 = 4 times 2 times 45 = 360${}^{4}\mathrm{C}_{3} \times {}^{2}\mathrm{C}_{1} \times {}^{10}\mathrm{C}_{8} = 4 \times 2 \times 45 = 360$ 2) 3 Engineers, 2 Doctors, 7 Professors: ^4mathrmC_3 times ^2mathrmC_2 times ^10mathrmC_7 = 4 times 1 times 120 = 480${}^{4}\mathrm{C}_{3} \times {}^{2}\mathrm{C}_{2} \times {}^{10}\mathrm{C}_{7} = 4 \times 1 \times 120 = 480$ 3) 4 Engineers, 1 Doctor, 7 Professors: ^4mathrmC_4 times ^2mathrmC_1 times ^10mathrmC_7 = 1 times 2 times 120 = 240${}^{4}\mathrm{C}_{4} \times {}^{2}\mathrm{C}_{1} \times {}^{10}\mathrm{C}_{7} = 1 \times 2 \times 120 = 240$ 4) 4 Engineers, 2 Doctors, 6 Professors: ^4mathrmC_4 times ^2mathrmC_2 times ^10mathrmC_6 = 1 times 1 times 210 = 210${}^{4}\mathrm{C}_{4} \times {}^{2}\mathrm{C}_{2} \times {}^{10}\mathrm{C}_{6} = 1 \times 1 \times 210 = 210$ ### Step 1: Calculate Final Probability Summing favorable cases: n(E) = 360 + 480 + 240 + 210 = 1290$n(E) = 360 + 480 + 240 + 210 = 1290$ textP(E) = frac12901820 = frac129182$\text{P}(E) = \frac{1290}{1820} = \frac{129}{182}$ ### Pattern Recognition When constraints involve 'at least' for multiple groups simultaneously with a large committee size, creating structured exhaustive case distributions is safer and less error-prone than standard bijection/subtraction methods. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 12 Mathematics: Probability