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:

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$- \frac{1}{8}$

$\frac{1}{8}$

$\frac{3}{16}$

$- \frac{3}{16}$

answer is B.

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Detailed Solution

The possibilities for getting one consecutive head are $\{H T T T T, T H T T T, T T H T T, T T T H T, T T T T H, H T H T T, H T T H T, H T T T H, T H T H T, T T H T H, T H T T H, H T H T H,\}$ 12 ways

The possibilities for getting two consecutive heads are

$\{H H T T T, T H H T T, T T H H T, T T T H H, H H T H T, H H T T H, T H H T H, H T H H T, H T T H H, T H T H H, H H T H H,\}$ 11 ways

The possibilities for getting three consecutive heads are

$\{H H H T T, T H H H T, T T H H H, H H H T H, H T H H H\}$ , 5 ways

The possibilities for getting four consecutive heads are

$\{H H H H T, T H H H H\}$ 2 ways

The possibilities for getting five consecutive heads is $\{H H H H H\}$

k = no. of times head occur consecutively

Now expectation

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