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

We begin by recalling the fundamental fact that tossing an unbiased coin five times produces $$2^{5}=32$$ equally likely sequences.

The random variable $$X$$ takes only four possible values:

$$X=5$$ if the sequence contains five consecutive heads,

$$X=4$$ if it contains four but not five consecutive heads,

$$X=3$$ if it contains exactly three consecutive heads and never four or five in a row,

$$X=-1$$ in every other case.

We now find the probability of each of these four mutually exclusive events.

1. Probability that $$X=5$$

Five consecutive heads can occur in only one way, namely the single sequence $$\text{HHHHH}$$. Thus

$$P(X=5)=\frac{1}{32}.$$

2. Probability that $$X=4$$

We need four heads in a row but not five. The block of four heads can start either at the first toss or at the second toss.

    • If it starts at toss 1 we have $$\text{HHHHT}.$$

    • If it starts at toss 2 we have $$\text{THHHH}.$$

There are exactly two such sequences and none of them equals the string of five heads already counted, so

$$P(X=4)=\frac{2}{32}=\frac{1}{16}.$$

3. Probability that $$X=3$$

We require a block of exactly three consecutive heads while forbidding any block of four or five. Let the block HHH start at position k.

    (i) k = 1  ⇒  pattern $$\text{HHH}x_{4}x_{5}.$$
To avoid four heads, $$x_{4}=\text{T}.$$ The fifth toss can be either H or T, giving the two strings $$\text{HHHTH},\; \text{HHHTT}.$$

    (ii) k = 2  ⇒  pattern $$x_{1}\text{HHH}x_{5}.$$
Here $$x_{1}=\text{T}$$ (otherwise we would have four heads) and $$x_{5}=\text{T}$$ for the same reason, yielding the single string $$\text{THHHT}.$$

    (iii) k = 3  ⇒  pattern $$x_{1}x_{2}\text{HHH}.$$
To avoid a block of four heads, $$x_{2}=\text{T}.$$ The first toss $$x_{1}$$ can be H or T, giving $$\text{HTHHH},\; \text{TTHHH}.$$

Counting them all, we have

$$5$$ sequences: \;$$$\text{HHHTH},\, \text{HHHTT},\, \text{THHHT},\, \text{HTHHH}$$$, $$\text{TTHHH}.$$

Thus

$$P(X=3)=\frac{5}{32}.$$

4. Probability that $$X=-1$$

This is simply the complement of the three cases already counted:

$$$P(X=-1)=1-\Bigl[\frac{1}{32}+\frac{2}{32}+\frac{5}{32}\Bigr] =1-\frac{8}{32} =\frac{24}{32} =\frac{3}{4}.$$$

5. Expected value of $$X$$

By definition, the expectation of a discrete random variable is $$E[X]=\sum x_{i}\,P(X=x_{i}).$$ Substituting our four values we get

$$$ \begin{aligned} E[X] &= 5\!\left(\frac{1}{32}\right) +4\!\left(\frac{2}{32}\right) +3\!\left(\frac{5}{32}\right) +(-1)\!\left(\frac{24}{32}\right) \\[6pt] &= \frac{5}{32}+\frac{8}{32}+\frac{15}{32}-\frac{24}{32} \\[6pt] &= \frac{28-24}{32} \\[6pt] &= \frac{4}{32} \\[6pt] &= \frac{1}{8}. \end{aligned} $$$

Hence, the correct answer is Option B.