ProbabilityHardBloom L3

Question

An unbiased coin is tossed 12 times. The probability that at least 7 consecutive heads show up is

Options

A.$\frac{5}{256}$
B.$\frac{1}{32}$
C.$\frac{3}{128}$
D.$\frac{7}{256}$

Solution

Sol. [Key Concept: Coin tossed 12 times: probability of the head-count event.]

$\underset{7}{\overset{HH - - H}{︸}}\underset{H\text{~}\text{or}\text{~}T}{\overset{XX - - - X}{︸}}$ $P = \frac{1}{2^{7}}$

T H H -----H X - - X $\mathbf{P} = \frac{1}{2^{8}}$

X T H -----H X - - X $\mathbf{P} = \frac{1}{2^{8}}$

⋮

$$\text{~}\text{X- - XT}\text{~}\underset{\text{last}\text{~}7}{\overset{HH - - H}{︸}}\ P = \frac{1}{2^{8}}$$

$$\begin{matrix} \Rightarrow \ P & \ = \frac{1}{2^{7}} + \underset{5\text{~}\text{times}\text{~}}{\overset{\frac{1}{2^{8}} + \frac{1}{2^{8}} + \frac{1}{2^{8}} + \ldots + \frac{1}{2^{8}}}{︸}} = \frac{1}{2^{7}} + \frac{5}{2^{8}} \\ & \ = \frac{7}{2^{8}} \end{matrix}$$

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