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JEE Main 2023 · Mathematics · Probability · hard

A fair coin is tossed repeatedly until two consecutive heads appear for the first time. Given that this happens within the first 6 tosses, the…

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A fair coin is tossed repeatedly until two consecutive heads appear for the first time. Given that this happens within the first 66 tosses, the probability that it happens exactly on the 6th6^{\text{th}} toss is
  1. A. 564\dfrac{5}{64}
  2. B. 543\dfrac{5}{43}
  3. C. 532\dfrac{5}{32}
  4. D. 843\dfrac{8}{43}
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Answer: (B) 543\dfrac{5}{43}

Count sequences in which \(HH\) first appears at the end of toss nn: the sequence must end in \(THH\) (for n≥3n\ge3) with no earlier \(HH\), and the counts follow the Fibonacci pattern 1,1,2,3,51,1,2,3,5 for n=2,3,4,5,6n=2,3,4,5,6. Their probabilities are cn2n\frac{c_n}{2^n}, so P(within 6)=14+18+216+332+564=16+8+8+6+564=4364P(\text{within }6)=\frac14+\frac18+\frac2{16}+\frac3{32}+\frac5{64}=\frac{16+8+8+6+5}{64}=\frac{43}{64}. Also P(exactly 6)=564P(\text{exactly }6)=\frac5{64}. The conditional probability is 5/6443/64=543\frac{5/64}{43/64}=\frac5{43}. A direct enumeration of all 262^6 outcomes confirms these counts.
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