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

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