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, otherwise X takes the value — 1. Then the expected value of X, is :

Question

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, otherwise X takes the value — 1. Then the expected value of X, is : (A) (1/8) (B) (3/16) (C) -(1/8) (D) -(3/16)

Tardigrade Admin · Accepted Answer

Expected value = ∑ XP(k) -(1/32)-(12/32)-(11/32)+(15/32)+(8/32)+(5/32) = (28-24/32) = (4/32) = (1/8)

Q. 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, otherwise X takes the value — 1. Then the expected value of X, is :

$\frac{1}{8}$

$\frac{3}{16}$

$- \frac{1}{8}$

$- \frac{3}{16}$

Expected value $= \sum XP (k)$
$- \frac{1}{32} - \frac{12}{32} - \frac{11}{32} + \frac{15}{32} + \frac{8}{32} + \frac{5}{32}$
$= \frac{28 - 24}{32} = \frac{4}{32} = \frac{1}{8}$

Q. 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, otherwise X takes the value — 1. Then the expected value of X, is :

81​

163​

−81​

−163​

Expected value =∑XP(k) −321​−3212​−3211​+3215​+328​+325​ =3228−24​=324​=81​

$\frac{1}{8}$

$\frac{3}{16}$

$- \frac{1}{8}$

$- \frac{3}{16}$

Expected value $= \sum XP (k)$
$- \frac{1}{32} - \frac{12}{32} - \frac{11}{32} + \frac{15}{32} + \frac{8}{32} + \frac{5}{32}$
$= \frac{28 - 24}{32} = \frac{4}{32} = \frac{1}{8}$