Q113 / 120 · NDA 2024 (Phase 2) · Mathematics

A fair coin is tossed till two heads occur in succession. What is the probability that the number of tosses required is less than 6 ?

  • A
    $\frac{5}{64}$
  • Correct answer
    B
    $\frac{15}{32}$
  • C
    $\frac{31}{64}$
  • D
    $\frac{19}{32}$

Solution

$\because$ Fair coin is tossed till two heads occur in succession. $\therefore \quad \mathrm{P}$ (the number of tosses required in less than 6 ) $=\mathrm{P}(\mathrm{HH})+\mathrm{P}(\mathrm{THH})+\mathrm{P}(\mathrm{TTHH})+\mathrm{P}(\mathrm{TTTHH})$ $=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5$ $\frac{=\left(\frac{1}{2}\right)^2\left[1-\left(\frac{1}{2}\right)^4\right]}{1-\frac{1}{2}}=\frac{\frac{1}{4}\left(1-\frac{1}{16}\right)}{\frac{1}{2}}=\frac{1}{2}\left(\frac{15}{16}\right)=\frac{15}{32}$

Hence, the correct answer is option: 'B'.

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