Question

If a fair coin is tossed $5$ times, the probability that heads does not occur two or more times in a row is

• A. $\frac{12}{2^5}$
• B. $\frac{13}{2^5}$
• C. $\frac{14}{2^5}$
• D. $\frac{15}{2^5}$

Answer 1

The correct option is B $\frac{13}{2^5}$

Probability that heads does not occurs two or more times in a row = Probability that head will not appears in any trial or only one head appears or heads will appear in alternative trial.
Therefore,
Case $\left(1\right):$ All tail appear
$p_1={\left(\frac{1}{2}\right)}^5=\frac{1}{2^5}$

Case $\left(2\right):$ Only one head appear i.e., $4T,1H$
$p_2={}^5{C}_4\times {\left(\frac{1}{2}\right)}^4\times \left(\frac{1}{2}\right)=\frac{5}{2^5}$

Case $\left(3\right):$ heads appear in alternate position $(3T,2H)$ i.e.,
$\left\{\_\_T\_\_T\_\_T\_\_\right\}{p}_3={}^4{C}_2\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^3=\frac{6}{2^5}$
where $\_\_$ represent places where head can occur.

Case $\left(4\right):$ heads appear in alternate position $(3H,2T)$ i.e., $\left\{\_\_T\_\_T\_\_\right\}$
$p_4={}^3{C}_3\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^3=\frac{1}{2^5}$

Hence, required probability$=\frac{1+5+6+1}{2^5}=\frac{13}{2^5}$