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

Number of ways 3 consecutive heads can appers <br><br>(1) HHHT_ <br><br>(2) _THHH <br><br>(3) THHHT <br><br>$\therefore$ Probablity of getting 3 consecutive heads <br><br>= ${2 \over {32}}$ + ${2 \over {32}}$ + ${1 \over {32}}$ = ${5 \over {32}}$ <br><br>Number of ways 4 consecutive heads can appers <br><br>(1) HHHHT <br><br>(2) THHHH <br><br>$\therefore$ Probablity of getting 4 consecutive heads <br><br>= ${1 \over {32}}$ + ${1 \over {32}}$ = ${2 \over {32}}$ <br><br>Number of ways 5 consecutive heads can appers <br><br>(1) HHHHH <br><br>$\therefore$ Probablity of getting 5 consecutive heads <br><br>= ${1 \over {32}}$ <br><br>Now Probablity of getting 0, 1, and 2 consecutive heads <br><br>= 1 - $\left( {{5 \over {32}} + {2 \over {32}} + {1 \over {32}}} \right)$ = ${{{24} \over {32}}}$ <br><br>Now, Expectation <br><br>= (-1) $\times$ ${{{24} \over {32}}}$ + 3 $\times$ ${{{5} \over {32}}}$ + 4 $\times$ ${{{2} \over {32}}}$ + 5 $\times$ ${{{1} \over {32}}}$ <br><br>= ${1 \over 8}$