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: $ 3 T$ and $2 H $ but not consecutive heads.
T T T (4 gaps) and place 2 Heads in gaps $ \Rightarrow { }^4 C _2=6$ ways
$\text { Case-IV: } 2 T \text { and } 3 H \text { but not consecutive heads. } $
$ T T (3 gaps ) \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 $=\frac{13}{256}$