### Related Formula S_infty = fraca1 - r$S_{\infty} = \frac{a}{1 - r}$ Where S_infty$S_{\infty}$ is the sum of an infinite geometric progression, a$a$ is the first term, and r$r$ is the common ratio. ### Core Logic Let success (S$S$) be rolling a 2, and failure (F$F$) be rolling anything else. P(S) = frac16$P(S) = \frac{1}{6}$ P(F) = frac56$P(F) = \frac{5}{6}$ We want the probability that the first success occurs on an even number of throws (2nd, 4th, 6th, dots$\dots$). The sequence of events for success on even throws is: - Success on 2nd throw: F, S$F, S$ - Success on 4th throw: F, F, F, S$F, F, F, S$ - Success on 6th throw: F, F, F, F, F, S$F, F, F, F, F, S$ Writing this as a sum of probabilities: textRequired Probability = P(F)P(S) + P(F)^3 P(S) + P(F)^5 P(S) + dots$\text{Required Probability} = P(F)P(S) + P(F)^3 P(S) + P(F)^5 P(S) + \dots$ = left(frac56right)left(frac16right) + left(frac56right)^3left(frac16right) + left(frac56right)^5left(frac16right) + dots$= \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right) + \dots$ ### Step 1: Compute Infinite Series Sum This is an infinite geometric series with: First term a = frac56 times frac16 = frac536$a = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ Common ratio r = left(frac56right)^2 = frac2536$r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$ Applying the sum formula: S_infty = fracfrac5361 - frac2536$S_{\infty} = \frac{\frac{5}{36}}{1 - \frac{25}{36}}$ = fracfrac536frac1136$= \frac{\frac{5}{36}}{\frac{11}{36}}$ = frac511$= \frac{5}{11}$ ### Pattern Recognition For alternating success/failure probabilities P(textEven) = fracq cdot p1 - q^2$P(\text{Even}) = \frac{q \cdot p}{1 - q^2}$ and P(textOdd) = fracp1 - q^2$P(\text{Odd}) = \frac{p}{1 - q^2}$. Knowing this format immediately turns it into a 5-second mental calculation: frac(5/6)(1/6)1 - 25/36 = 5/11$\frac{(5/6)(1/6)}{1 - 25/36} = 5/11$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability Class 11 Mathematics: Sequences and Series

### Related Formula textBayes' Theorem: P(E_1 | E) = fracP(E_1)P(E | E_1)P(E_1)P(E | E_1) + P(E_2)P(E | E_2)$\text{Bayes' Theorem: } P(E_1 | E) = \frac{P(E_1)P(E | E_1)}{P(E_1)P(E | E_1) + P(E_2)P(E | E_2)}$ ### Core Logic Let E_1$E_1$ be the event that Bag A is selected, and E_2$E_2$ be the event that Bag B is selected. P(E_1) = P(E_2) = frac12$P(E_1) = P(E_2) = \frac{1}{2}$ Let E$E$ be the event that a white ball is drawn. From Bag A (3 white, 7 red, total 10): P(E | E_1) = frac310$P(E | E_1) = \frac{3}{10}$ From Bag B (3 white, 2 red, total 5): P(E | E_2) = frac35$P(E | E_2) = \frac{3}{5}$ ### Step 1: Calculating the Target Probability We need to find the probability that the ball was drawn from Bag A given it is white, i.e., P(E_1 | E)$P(E_1 | E)$. P(E_1 | E) = fracfrac12 times frac310frac12 times frac310 + frac12 times frac35$P(E_1 | E) = \frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5}}$ Canceling out frac12$\frac{1}{2}$ from the numerator and the denominator: P(E_1 | E) = fracfrac310frac310 + frac610 = frac33 + 6 = frac39 = frac13$P(E_1 | E) = \frac{\frac{3}{10}}{\frac{3}{10} + \frac{6}{10}} = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}$ ### Pattern Recognition Reverse probability with disjoint prior states directly signals Bayes' theorem. Canceling prior probability terms (P(E_1)=P(E_2)$P(E_1)=P(E_2)$) speeds up the calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Related Formula P(X=k) = ^nC_k cdot p^k cdot q^n-k$P(X=k) = ^nC_k \cdot p^k \cdot q^{n-k}$ ### Core Logic Given P(H) = 2P(T)$P(H) = 2P(T)$. Since P(H) + P(T) = 1$P(H) + P(T) = 1$, we get 2P(T) + P(T) = 1 implies 3P(T) = 1 implies P(T) = frac13$2P(T) + P(T) = 1 \implies 3P(T) = 1 \implies P(T) = \frac{1}{3}$. Then, P(H) = frac23$P(H) = \frac{2}{3}$. The coin is tossed 3 times. We need the probability of getting exactly 2 tails and 1 head. Using binomial probability: P(2T, 1H) = ^3C_2 times (P(T))^2 times (P(H))^1$P(2T, 1H) = ^3C_2 \times (P(T))^2 \times (P(H))^1$ = 3 times left(frac13right)^2 times left(frac23right)$= 3 \times \left(\frac{1}{3}\right)^2 \times \left(\frac{2}{3}\right)$ = 3 times frac19 times frac23 = frac627 = frac29$= 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

### Core Logic Total marbles = 10 + 30 + 20 + 15 = 75$= 10 + 30 + 20 + 15 = 75$. Drawings are made with replacement, so the events are independent. ### Step 1: Probability Calculation Probability of first drawing a red marble: P(R) = frac1075$P(R) = \frac{10}{75}$. Probability of second drawing a white marble: P(W) = frac3075$P(W) = \frac{30}{75}$. Since they are independent: P(R text and W) = frac1075 times frac3075 = frac475$P(R \text{ and } W) = \frac{10}{75} \times \frac{30}{75} = \frac{4}{75}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability