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  4. A normal coin is tossed six times. If p =( a / b ) where a , b are relatively prime positive integers, denotes the chance that there will be exactly one sequence of 3 consecutive heads in six tosses, then find the value of ((a+b+1/4)).

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Q. A normal coin is tossed six times. If $p =\frac{ a }{ b }$ where $a , b$ are relatively prime positive integers, denotes the chance that there will be exactly one sequence of 3 consecutive heads in six tosses, then find the value of $\left(\frac{a+b+1}{4}\right)$.

Probability - Part 2

Solution:

H H H $\times x \quad x \rightarrow H$ or $T$
$\text { THHHT } \times $
$\times \text { THHHT } $
$\times \times \text { THHH }$
$\therefore P(E)=\frac{1}{16}+\frac{1}{32}+\frac{1}{32}+\frac{1}{16}=\frac{2}{16}+\frac{1}{16}=\frac{3}{16}=\frac{a}{b} \Rightarrow\left(\frac{a+b+1}{4}\right)=5$