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A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Solution & Explanation

### Related Formula S_infty = fraca1 - r Where S_infty is the sum of an infinite geometric progression, a is the first term, and r is the common ratio. ### Core Logic Let success (S) be rolling a 2, and failure (F) be rolling anything else. P(S) = frac16 P(F) = frac56 We want the probability that the first success occurs on an even number of throws (2nd, 4th, 6th, dots). The sequence of events for success on even throws is: - Success on 2nd throw: F, S - Success on 4th throw: F, F, F, S - Success on 6th throw: 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 = left(frac56right)left(frac16right) + left(frac56right)^3left(frac16right) + left(frac56right)^5left(frac16right) + dots ### Step 1: Compute Infinite Series Sum This is an infinite geometric series with: First term a = frac56 times frac16 = frac536 Common ratio r = left(frac56right)^2 = frac2536 Applying the sum formula: S_infty = fracfrac5361 - frac2536 = fracfrac536frac1136 = frac511 ### Pattern Recognition For alternating success/failure probabilities P(textEven) = fracq cdot p1 - q^2 and P(textOdd) = fracp1 - q^2. Knowing this format immediately turns it into a 5-second mental calculation: 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

Reference Study Guides

Probability

More Probability Previous-Year Questions — Page 5

Q11 jee_main_2024_30_january_evening Bayes Theorem
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is :
  • A. frac14
  • B. frac19
  • C. frac13
  • D. frac310

Solution

### 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) ### Core Logic Let E_1 be the event that Bag A is selected, and E_2 be the event that Bag B is selected. P(E_1) = P(E_2) = frac12 Let E be the event that a white ball is drawn. From Bag A (3 white, 7 red, total 10): P(E | E_1) = frac310 From Bag B (3 white, 2 red, total 5): P(E | E_2) = frac35 ### 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) = fracfrac12 times frac310frac12 times frac310 + frac12 times frac35 Canceling out frac12 from the numerator and the denominator: P(E_1 | E) = fracfrac310frac310 + frac610 = frac33 + 6 = frac39 = frac13 ### Pattern Recognition Reverse probability with disjoint prior states directly signals Bayes' theorem. Canceling prior probability terms (P(E_1)=P(E_2)) speeds up the calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability
Q11 jee_main_2024_30_jan_morning Classical Probability
Two integers x and y are chosen with replacement from the set \0, 1, 2, 3, dots, 10\. Then the probability that |x - y| > 5 is:
  • A. frac30121
  • B. frac62121
  • C. frac60121
  • D. frac31121

Solution

### Related Formula P(E) = fractextNumber of favorable outcomestextTotal number of possible outcomes ### Core Logic Total possible selections for (x, y) with replacement from 0, 1, dots, 10 is 11 times 11 = 121. We need pairs (x, y) such that |x - y| > 5, which means x - y > 5 or y - x > 5. ### Step 1: Counting Favorable Cases Assume x < y, so we need y - x geq 6. If x = 0 Rightarrow y in \6, 7, 8, 9, 10\ (5 ways) If x = 1 Rightarrow y in \7, 8, 9, 10\ (4 ways) If x = 2 Rightarrow y in \8, 9, 10\ (3 ways) If x = 3 Rightarrow y in \9, 10\ (2 ways) If x = 4 Rightarrow y in \10\ (1 way) If x ge 5, there are no possible values for y strictly greater than x satisfying the condition. ### Step 2: Total Probability The number of cases for y > x is 5 + 4 + 3 + 2 + 1 = 15. By symmetry, the number of cases for x > y is also 15. Total favorable cases = 15 times 2 = 30. Required probability = frac30121. ### Pattern Recognition Absolute difference conditions |x - y| > k on discrete sets cleanly split into symmetric additive series 1+2+...+n. Calculate one half and multiply by 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability
Q18 jee_main_2024_31_jan_evening Binomial Distribution / Independent Events
A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-
  • A. frac29
  • B. frac19
  • C. frac227
  • D. frac127

Solution

### Related Formula P(X=k) = ^nC_k cdot p^k cdot q^n-k ### Core Logic Given P(H) = 2P(T). Since P(H) + P(T) = 1, we get 2P(T) + P(T) = 1 implies 3P(T) = 1 implies P(T) = frac13. Then, P(H) = frac23. 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 = 3 times left(frac13right)^2 times left(frac23right) = 3 times frac19 times frac23 = frac627 = frac29 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability
Q16 jee_main_2024_31_jan_morning Independent Events
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is
  • A. frac225
  • B. frac425
  • C. frac23
  • D. frac475

Solution

### Core Logic Total marbles = 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. Probability of second drawing a white marble: P(W) = frac3075. Since they are independent: P(R text and W) = frac1075 times frac3075 = frac475. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability

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