Question

A normal coin tossed repeatedly until two consecutive heads is obtained. the probability two consecutive heads occurs on the seventh and eight flips is equal to

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

Answer 1

Answer: (C)

$\frac{13}{256}$

We know that $7^{th}$ and $8^{th}$ tips have to be heads and also the $6^{th}$ flip should be tails. If $6^{th}$ flip is heads, $7^{th}$ flip is also heads , so there want be need for $8^{th}$ flip. So, $6^{th}$ flip should be tails.

$\begin{array}{c}6thflip\to tails\\ 7thflip\to heads\\ 8thflip\to heads\end{array}\}$ $\to$ Probability $=\frac{1}{2^3}$

We, have to decide the outcomes of remaining $5$ tails such that there are no consecutive heads.

Case i) Only $1$ heads in $5$ trails

no. of ways $\Rightarrow 5_9\Rightarrow 5$ Probability = $\frac{5}{25}$

case ii) No heads in $5$ trails

no of ways $\Rightarrow 1$ Probability = $\frac{1}{25}$

cosec iii) $2$ heads in $5$ trails

$-T-T-T-$

we have to arrange $2$ heads in $4$ block

in ${}^4{C}_2$ ways $\Rightarrow 6$ Probability $=\frac{6}{25}$

cosec iv) $3$ heads :-

no of ways = $1$ $\underset{–}{H}\underset{–}{T}\underset{–}{H}\underset{–}{T}\underset{–}{H}$ prob $=\frac{1}{2^5}$

$\therefore$ Probability $\Rightarrow \frac{1}{2^3}[\frac{5}{2^5}+\frac{1}{2^5}+\frac{6}{2^5}+\frac{1}{2^5}]$

$\Rightarrow \frac{13}{2^5}=\frac{13}{256}$

probability that $2$ head occur on $7^{th}$ and $8^{th}$ trails is $\frac{13}{256}$