An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, other wise X takes the value -1.The expected value of X, is:

The possibilities for getting one consecutive head are $\left\{HTTTT, THTTT, TTHTT, TTTHT,TTTTH, HTHTT, HTTHT,HTTTH,THTHT,TTHTH,THTTH,HTHTH,\right\}$ 12 ways

The possibilities for getting two consecutive heads are 

$\left\{HHTTT,THHTT,TTHHT,TTTHH,HHTHT,HHTTH,THHTH,HTHHT,HTTHH,THTHH, HHTHH,\right\}$ 11 ways

The possibilities for getting three consecutive heads are 

$\left\{HHHTT,THHHT,TTHHH,HHHTH,HTHHH\right\}$, 5 ways 

The possibilities for getting four consecutive heads are 

$\left\{HHHHT, THHHH\right\}$ 2 ways 

The possibilities for getting five consecutive heads is $\left\{HHHHH\right\}$

k = no. of times head occur consecutively

 Now expectation 

$=\sum XP\left(k\right)=\left(-1\right)\times \frac{1}{32}+\left(-1\right)\times \frac{12}{32}+\left(-1\right)\times \frac{11}{32}+3\times \frac{5}{32}+4\times \frac{2}{32}+5\times \frac{1}{32}=+\frac{1}{8}$