### Related Formula Bayes' Theorem for conditional probability is formulated as: P(E_1|H) = fracP(E_1) cdot P(H|E_1)P(E_1) cdot P(H|E_1) + P(E_2) cdot P(H|E_2)$P(E_1|H) = \frac{P(E_1) \cdot P(H|E_1)}{P(E_1) \cdot P(H|E_1) + P(E_2) \cdot P(H|E_2)}$ ### Core Logic Let the events be: E_1$E_1$: Selection of an unbiased coin. E_2$E_2$: Selection of the two-headed (biased) coin. H$H$: Head turns up on the toss. Syllabus values: P(E_1) = frac1920, quad P(E_2) = frac120$P(E_1) = \frac{19}{20}, \quad P(E_2) = \frac{1}{20}$ P(H|E_1) = frac12, quad P(H|E_2) = 1$P(H|E_1) = \frac{1}{2}, \quad P(H|E_2) = 1$ ### Step 1: Total Probability Calculation The overall probability of obtaining a head is: P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2)$P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2)$ P(H) = frac1920 cdot frac12 + frac120 cdot 1 = frac1940 + frac240 = frac2140$P(H) = \frac{19}{20} \cdot \frac{1}{2} + \frac{1}{20} \cdot 1 = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$ ### Step 2: Apply Bayes Theorem We need the probability that the coin is unbiased given a head showed up: P(E_1|H) = fracfrac1940frac2140 = frac1921$P(E_1|H) = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}$ Thus, fracmn = frac1921 implies m = 19, n = 21$\frac{m}{n} = \frac{19}{21} \implies m = 19, n = 21$ since gcd(19, 21) = 1$\gcd(19, 21) = 1$. ### Step 3: Evaluate final expression Calculate n^2 - m^2$n^2 - m^2$: n^2 - m^2 = 21^2 - 19^2 = 441 - 361 = 80$n^2 - m^2 = 21^2 - 19^2 = 441 - 361 = 80$ ### Pattern Recognition Bayes' Theorem split problems are easily handled by constructing paths: textunbiased path = 19 times 1 = 19$\text{unbiased path} = 19 \times 1 = 19$, textbiased path = 1 times 2 = 2$\text{biased path} = 1 \times 2 = 2$. Probability = frac1919+2 = frac1921$\frac{19}{19+2} = \frac{19}{21}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula The sum of all probabilities in a probability distribution is strictly equal to 1$1$: sum P(X_i) = 1$\sum P(X_i) = 1$ ### Core Logic Let P(X=2) = P(X=3) = q$P(X=2) = P(X=3) = q$. From the total probability rule: P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ p + p + q + q = 1 implies 2p + 2q = 1 implies p + q = frac12 quad dots text(i)$p + p + q + q = 1 \implies 2p + 2q = 1 \implies p + q = \frac{1}{2} \quad \dots \text{(i)}$ ### Step 1: Computing Expectations Compute E(X)$E(X)$: E(X) = 0cdot p + 1cdot p + 2cdot q + 3cdot q = p + 5q$E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q$ Compute E(X^2)$E(X^2)$: E(X^2) = 0^2cdot p + 1^2cdot p + 2^2cdot q + 3^2cdot q = p + 13q$E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q$ ### Step 2: Solve the Linear System Given E(X^2) = 2E(X)$E(X^2) = 2E(X)$: p + 13q = 2(p + 5q)$p + 13q = 2(p + 5q)$ p + 13q = 2p + 10q implies p = 3q$p + 13q = 2p + 10q \implies p = 3q$ Substitute p = 3q$p = 3q$ into equation (i): 3q + q = frac12 implies 4q = frac12 implies q = frac18$3q + q = \frac{1}{2} \implies 4q = \frac{1}{2} \implies q = \frac{1}{8}$ Then, p = 3left(frac18right) = frac38$p = 3\left(\frac{1}{8}\right) = \frac{3}{8}$. ### Step 3: Calculate 8p - 1 Now we evaluate the required expression: 8p - 1 = 8left(frac38right) - 1 = 3 - 1 = 2$8p - 1 = 8\left(\frac{3}{8}\right) - 1 = 3 - 1 = 2$ ### Pattern Recognition Always combine basic distribution axioms (sum of probabilities = 1) with structural definition equations (E(X) = sum x P(x)$E(X) = \sum x P(x)$) to systematically eliminate unknown parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

### Related Formula A 2 times 2$2 \times 2$ matrix A = beginpmatrix a & b \\ c & d endpmatrix$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is invertible if and only if its determinant is non-zero: det(A) = ad - bc neq 0$\det(A) = ad - bc \neq 0$ ### Step 1: Count Total Matrix Sample Space Each of the 4 entry slots in the 2 times 2$2 \times 2$ matrix has 2 binary choices (0 or 1) : textTotal Matrices = 2^4 = 16$\text{Total Matrices} = 2^4 = 16$ ### Step 2: Count Favorable Non-Zero Determinant Matrices Since elements are 0 or 1, the products ad$ad$ and bc$bc$ can only evaluate to 0 or 1. For ad - bc neq 0$ad - bc \neq 0$, we have two distinct cases [cite: 4051, 4054]: - **Case I:** ad = 1$ad = 1$ and bc = 0$bc = 0$ . ad = 1 Rightarrow a = 1, d = 1$ad = 1 \Rightarrow a = 1, d = 1$ (1 configuration). bc = 0 Rightarrow (b, c) in \(0,0), (0,1), (1,0)\$bc = 0 \Rightarrow (b, c) \in \{(0,0), (0,1), (1,0)\}$ (3 configurations). textWays = 1 times 3 = 3 text matrices$\text{Ways} = 1 \times 3 = 3 \text{ matrices}$ - **Case II:** ad = 0$ad = 0$ and bc = 1$bc = 1$ . bc = 1 Rightarrow b = 1, c = 1$bc = 1 \Rightarrow b = 1, c = 1$ (1 configuration). ad = 0 Rightarrow (a, d) in \(0,0), (0,1), (1,0)\$ad = 0 \Rightarrow (a, d) \in \{(0,0), (0,1), (1,0)\}$ (3 configurations). textWays = 1 times 3 = 3 text matrices$\text{Ways} = 1 \times 3 = 3 \text{ matrices}$ textTotal Favorable Matrices = 3 + 3 = 6$\text{Total Favorable Matrices} = 3 + 3 = 6$ ### Step 3: Calculate Probability Divide the favorable count by the total sample size : P(E) = frac616 = frac38$P(E) = \frac{6}{16} = \frac{3}{8}$ ### Pattern Recognition For low-order matrix configuration spaces with binary inputs, directly analyzing the product outcomes (1-0=1$1-0=1$ or 0-1=-1$0-1=-1$) prevents long manual lists of all 16 matrices. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 12 Mathematics: Matrices and Determinants

### Related Formula For alternating multi-stage games continuing indefinitely, the total probability of winning is modeled as an infinite geometric series summation: S_infty = fraca1 - r$S_{\infty} = \frac{a}{1 - r}$ ### Core Logic First, analyze the total sample outcomes for a pair of standard dice (n(S) = 36$n(S) = 36$): - Outcomes giving a sum of 5: (1,4), (2,3), (3,2), (4,1) implies 4$(1,4), (2,3), (3,2), (4,1) \implies 4$ outcomes. p(A) = frac436 = frac19 implies p(A') = 1 - frac19 = frac89$p(A) = \frac{4}{36} = \frac{1}{9} \implies p(A') = 1 - \frac{1}{9} = \frac{8}{9}$ - Outcomes giving a sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) implies 5$(2,6), (3,5), (4,4), (5,3), (6,2) \implies 5$ outcomes. p(B) = frac536 implies p(B') = 1 - frac536 = frac3136$p(B) = \frac{5}{36} \implies p(B') = 1 - \frac{5}{36} = \frac{31}{36}$ ### Step 1: Setup Infinite Game Series Path For player A$A$ to win on the first throw, third throw, fifth throw, etc., the probability sequence expands as follows: P(textA wins) = p(A) + p(A') cdot p(B') cdot p(A) + [p(A') cdot p(B')]^2 cdot p(A) + dots infty$P(\text{A wins}) = p(A) + p(A') \cdot p(B') \cdot p(A) + [p(A') \cdot p(B')]^2 \cdot p(A) + \dots \infty$ This is a geometric progression where the first term a = p(A) = frac19$a = p(A) = \frac{1}{9}$ and the common ratio is: r = p(A') cdot p(B') = frac89 cdot frac3136 = frac6281$r = p(A') \cdot p(B') = \frac{8}{9} \cdot \frac{31}{36} = \frac{62}{81}$ ### Step 2: Calculate Invariant Sum Apply the infinite GP sum formula directly: P(textA wins) = fracfrac191 - frac6281 = fracfrac19frac1981 = frac19 cdot frac8119 = frac919$P(\text{A wins}) = \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{\frac{1}{9}}{\frac{19}{81}} = \frac{1}{9} \cdot \frac{81}{19} = \frac{9}{19}$ ### Pattern Recognition In infinite alternating games, player A's net probability can always be abbreviated shortcut-style to fracp_11 - (1-p_1)(1-p_2)$\frac{p_1}{1 - (1-p_1)(1-p_2)}$, where p_1$p_1$ is A's success probability and p_2$p_2$ is B's success probability. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability