ProbabilityHardBloom L3
Question
A fair coin is tossed ( $2m + 1$ ) times, the probability of getting at least ' $m$ ' consecutive heads is
Options
A.$\frac{m + 1}{2^{m}}$
B.$\frac{m + 3}{2^{m + 1}}$
C.$\frac{(m + 1)2^{m} - 1}{2^{2\text{ }m + 1}}$
D.$\frac{(m + 3)2^{m} - 1}{2^{2m + 1}}$
Solution
Sol. [Key Concept: Fair coin $(2m + 1)$ tosses: $P(more\ heads\ than\ tails) = \frac{1}{2}$.]
$$\underset{m}{\overset{HHH - HXX - - X}{︸}}\ P = \frac{1}{2^{m}} + \frac{m + 1}{2^{m + 1}} - \frac{1}{2}$$
$$\begin{matrix} & \\ \text{~}\text{THH--HXX--X}\text{~} & = \frac{(m + 3)2^{m} - 1}{2^{2m + 1}} \\ \text{~}\text{XTHH---HX---X}\text{~} & \\ \vdots & \end{matrix}$$
XX --ХТНН-Н
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