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Probability question

2023 · Shift 2 · Q19
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  5. /2023 · Shift 2 · Q19

Probability question

2023 · Shift 2 · Q19

JEE AdvancedMathematicsProbabilityMCQ+3 / −1
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}31​, then the probability that the experiment stops with head is :
  1. A
    13\frac{1}{3}31​
  2. B
    521\frac{5}{21}215​
  3. C
    421\frac{4}{21}214​
  4. D
    27\frac{2}{7}72​
View written solutionFree

Correct answer: B

  1. Let P(H)=13P(H)=\frac{1}{3}P(H)=31​ and P(T)=23P(T)=\frac{2}{3}P(T)=32​.

  2. The experiment stops when two consecutive tosses are the same. So the experiment can stop with head only if the last two tosses are HHHHHH.

  3. To stop with head, the sequence must alternate until the final HHHHHH appears. Hence possible sequences are: HH,THH,HTHH,THTHH,HTHTHH,…HH,\quad THH,\quad HTHH,\quad THTHH,\quad HTHTHH,\dotsHH,THH,HTHH,THTHH,HTHTHH,…

    These are of two types:

    • Starting directly with HHH: HH,  HTHH,  HTHTHH,…H H,\; H T H H,\; H T H T H H,\dotsHH,HTHH,HTHTHH,…
    • Starting with TTT: THH,  THTHH,  THTHTHH,…T H H,\; T H T H H,\; T H T H T H H,\dotsTHH,THTHH,THTHTHH,…

  4. Compute the probability of each type.

    Type 1: H(TH)nHH(TH)^nHH(TH)nH

    For n=0,1,2,…n=0,1,2,\dotsn=0,1,2,… P(H(TH)nH)=(13)n+2(23)nP\big(H(TH)^nH\big)=\left(\frac{1}{3}\right)^{n+2}\left(\frac{2}{3}\right)^nP(H(TH)nH)=(31​)n+2(32​)n

    So the total probability is

    =\frac{1}{9}\sum_{n=0}^\infty \left(\frac{2}{9}\right)^n =\frac{1}{9}\cdot \frac{1}{1-\frac{2}{9}} =\frac{1}{9}\cdot \frac{9}{7} =\frac{1}{7}.$$ ### Type 2: $T(H T)^nHH$ For $n=0,1,2,\dots$ $$P\big(T(HT)^nHH\big)=\left(\frac{2}{3}\right)^{n+1}\left(\frac{1}{3}\right)^{n+2}$$ So the total probability is $$S_2=\sum_{n=0}^\infty \left(\frac{2}{3}\right)^{n+1}\left(\frac{1}{3}\right)^{n+2} =\frac{2}{9}\sum_{n=0}^\infty \left(\frac{2}{9}\right)^n \cdot \frac{1}{3} =\sum_{n=0}^\infty \frac{2^{n+1}}{3^{2n+3}}.$$ Simplifying directly, $$S_2=\frac{2}{27}\sum_{n=0}^\infty \left(\frac{2}{9}\right)^n =\frac{2}{27}\cdot \frac{1}{1-\frac{2}{9}} =\frac{2}{27}\cdot \frac{9}{7} =\frac{2}{21}.$$

  5. Therefore, the required probability is S1+S2=17+221=321+221=521.S_1+S_2=\frac{1}{7}+\frac{2}{21}=\frac{3}{21}+\frac{2}{21}=\frac{5}{21}.S1​+S2​=71​+212​=213​+212​=215​.

  6. Hence the correct option is: 521\boxed{\frac{5}{21}}215​​ which is Option B.

  7. Comparison with stored answer: Stored correct answer is B, which matches our result.

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