Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is 29/45, then n is equal to:

Solution & Explanation

### Related Formula Total Probability Law: P(W) = P(W|B_1)P(B_1) + P(W|B_2)P(B_2) ### Core Logic Bag 1 contents: \4W, 5B\ (Total 9 balls). Bag 2 contents initially: \nW, 3B\ (Total n+3 balls). Case 1: Transferred ball is white (P = frac49): Bag 2 now has (n+1)W and 3B (Total n+4). Probability of drawing white = fracn+1n+4. Case 2: Transferred ball is black (P = frac59): Bag 2 now has nW and 4B (Total n+4). Probability of drawing white = fracnn+4. ### Step 1: Set up Equation and Solve Aggregate components via Total Probability Formula: left(frac49 times fracn+1n+4right) + left(frac59 times fracnn+4right) = frac2945 frac4(n+1) + 5n9(n+4) = frac2945 frac9n + 4n+4 = frac295 5(9n + 4) = 29(n + 4) 45n + 20 = 29n + 116 16n = 96 implies n = 6 ### Pattern Recognition Notice how the denominators inside conditional stages match up identically (n+4). Clear constants before running fraction line conversions to speed up single-variable systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability

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Q19 jee_main_2024_31_jan_morning Variance of Random Variable
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable X to be the number of rotten apples in a draw of two apples, the variance of X is
  • A. frac37153
  • B. frac57153
  • C. frac47153
  • D. frac40153

Solution

### Core Logic Total apples = 18 (3 rotten, 15 good). Random variable X = \0, 1, 2\ representing the number of rotten apples. ### Step 1: Probability Distribution P(X = 0) = frac^15C_2^18C_2 = frac105153 P(X = 1) = frac^3C_1 times ^15C_1^18C_2 = frac45153 P(X = 2) = frac^3C_2^18C_2 = frac3153 ### Step 2: Expectation E(X) = 0 times frac105153 + 1 times frac45153 + 2 times frac3153 = frac51153 = frac13 ### Step 3: Variance E(X^2) = 0 times frac105153 + 1 times frac45153 + 4 times frac3153 = frac57153 Var(X) = E(X^2) - (E(X))^2 = frac57153 - left(frac13right)^2 = frac57153 - frac17153 = frac40153 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

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