JEE Advanced 2023 Paper 2 Q02 Mathematics P&C and Probability Probability Hard

JEE Advanced 2023 Paper 2 · Q02 · Probability

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $1/3$, then the probability that the experiment stops with head is

  1. A. $1/3$
  2. B. $5/21$
  3. C. $4/21$
  4. D. $2/7$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Let $P_H$ denote stopping with head. The favorable sequences are those ending in $\ldots HH$. Cases: starts $H$ then immediately $H$: prob $(1/3)(1/3) = 1/9$. Or starts $H$ then alternates $T,H,T,H,\ldots$ before hitting $HH$. Or starts $T$ then transitions. By summing the geometric series for both starting outcomes and combining: $P(\text{stop with H}) = \sum_{k=0}^{\infty}\left[\left(\dfrac{1}{3}\right)\left(\dfrac{2}{3}\cdot\dfrac{1}{3}\right)^k\right]\cdot\dfrac{1}{3} + \sum_{k=0}^{\infty}\left[\left(\dfrac{2}{3}\cdot\dfrac{1}{3}\right)^{k+1}\right]\cdot\dfrac{1}{3}$. Simplifying: $P = (1/9)/(1 - 2/9) + (2/9)\cdot(1/3)/(1 - 2/9) = (1/9 + 2/27)/(7/9) = (5/27)\cdot(9/7) = 5/21$.

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