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Mathematics · Probability

JEE Advanced 2023 — Paper 2 — Question 2

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 13\frac{1}{3}, then the probability that the experiment stops with head is

  1. Option A:

    13\frac{1}{3}

  2. Option B:

    521\frac{5}{21}

    Correct
  3. Option C:

    421\frac{4}{21}

  4. Option D:

    27\frac{2}{7}0

Answer: B

Step-by-step solution

P(H)=13,P( T)=23P=P(H H )+{P( T H H )+P( T H T H H )+P( T H T H T H H )+…∞)+{P( H T H H )+P( H T H T H H )+P( H T H T H T H H )+..∞)=19+(23×19+23×13×23×19+…..∞)+(13×23×19+13×23×13×23×19+…∞)=19+2271−29+13×23×191−29=19+23×7+27×9=521.\begin{aligned} \mathrm{P}(\mathrm{H}) & =\frac{1}{3}, \mathrm{P}(\mathrm{~T})=\frac{2}{3} \\ \mathrm{P}= & \mathrm{P}(\mathrm{H} \text { H })+\{\mathrm{P}(\text { T H H })+\mathrm{P}(\text { T H T H H })+\mathrm{P}(\text { T H T H T H H })+\ldots \infty) \\ & +\{\mathrm{P}(\text { H T H H })+\mathrm{P}(\text { H T H T H H })+\mathrm{P}(\text { H T H T H T H H })+. . \infty) \\ & =\frac{1}{9}+\left(\frac{2}{3} \times \frac{1}{9}+\frac{2}{3} \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{9}+\ldots . . \infty\right)+\left(\frac{1}{3} \times \frac{2}{3} \times \frac{1}{9}+\frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{9}+\ldots \infty\right) \\ & =\frac{1}{9}+\frac{\frac{2}{27}}{1-\frac{2}{9}}+\frac{\frac{1}{3} \times \frac{2}{3} \times \frac{1}{9}}{1-\frac{2}{9}} \\ = & \frac{1}{9}+\frac{2}{3 \times 7}+\frac{2}{7 \times 9}=\frac{5}{21} . \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Mathematics
Chapter
Probability
Topic
Introduction to Probability

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