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)).

Question

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)). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H H H × x x arrow H or T text THHHT × × text THHHT × × text THHH ∴ P(E)=(1/16)+(1/32)+(1/32)+(1/16)=(2/16)+(1/16)=(3/16)=(a/b) ⇒((a+b+1/4))=5

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 $(\frac{a + b + 1}{4})$ .

H H H $\times x x \to H$ or $T$
$THHHT \times$
$\times THHHT$
$\times \times THHH$
$∴ P (E) = \frac{1}{16} + \frac{1}{32} + \frac{1}{32} + \frac{1}{16} = \frac{2}{16} + \frac{1}{16} = \frac{3}{16} = \frac{a}{b} \Rightarrow (\frac{a + b + 1}{4}) = 5$

Q. A normal coin is tossed six times. If p=ba​ 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 (4a+b+1​).

H H H ×xx→H or T THHHT × × THHHT ×× THHH ∴P(E)=161​+321​+321​+161​=162​+161​=163​=ba​⇒(4a+b+1​)=5

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 $(\frac{a + b + 1}{4})$ .

H H H $\times x x \to H$ or $T$
$THHHT \times$
$\times THHHT$
$\times \times THHH$
$∴ P (E) = \frac{1}{16} + \frac{1}{32} + \frac{1}{32} + \frac{1}{16} = \frac{2}{16} + \frac{1}{16} = \frac{3}{16} = \frac{a}{b} \Rightarrow (\frac{a + b + 1}{4}) = 5$