Question

A fair coin is tossed repeatedly until two consecutive heads is obtained. The probability that two consecutive heads occur on the seventh and eights flips is equal to

• A. $\frac{11}{256}$
• B. $\frac{15}{256}$
• C. $\frac{13}{256}$
• D. $\frac{17}{256}$

Answer 1

Answer: (C)

$\times \times \times \times \times THH$
Now at $6^{\text{th }}$ place Tail must be present. (think!)
Possible cases in first 5 places.
Case-I: T T T T T All tails $=1$ way;
Case-II: 4 T and H i.e. T T T T H $\Rightarrow \frac{5!}{4!}=5$ ways
Case-III: $3T$ and $2H$ but not consecutive heads.
T T T (4 gaps) and place 2 Heads in gaps $\Rightarrow {}^4{C}_2=6$ ways
$\text{ Case-IV: }2T\text{ and }3H\text{ but not consecutive heads. }$
$TT(3gaps)\text{ and places }3\text{ Heads in gaps. }\Rightarrow {}^3{C}_3=1\text{ ways }$
$\Rightarrow$ Total 13 cases and in every case probability is $\frac{1}{256}$.
$\therefore$ Required probability $=\frac{13}{256}$