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) $\dfrac{1}{8}$
B) $-\dfrac{1}{8}$
C) $\dfrac{3}{16}$
D) $-\dfrac{3}{16}$

Answer 1

Hint: First find the probability of getting a number of heads simultaneously by substituting the value of k from 1 to 5. After that use the formula of the expected value of X, \[\sum{XP\left( k \right)}\]. For k = 1, 2, 3, put $X=-1$ and for other values of k, put $X=k$. Then solve it to get the desired result.

Complete step by step answer:
The total number of outcomes for the event is ${{2}^{5}}=32$.
Find the probability for getting number of heads occurring simultaneously,
For k = 0, the probable outcome will be,
$E\left( k \right)=\left\{ TTTTT \right\}=1$
$P\left( k \right)=\dfrac{1}{32}$
For k = 1, the probable outcome will be,
\[E\left( k \right)=\left\{ \begin{align}
  & HTTTT,THTTT,TTHTT,TTTHT,TTTTH,HTHTT \\
 & HTTHT,HTTTH,THTHT,THTTH,TTHTH,HTHTH \\
\end{align} \right\}=12\]
$\Rightarrow P\left( k \right)=\dfrac{12}{32}$
For k = 2, the probable outcome will be,
\[E\left( k \right)=\left\{ \begin{align}
  & HHTTT,THHTT,TTHHT,TTTHH,HHTHT,HHTTH \\
 & THHTH,HTHHT,HTTHH,THTHH,HHTHH \\
\end{align} \right\}=11\]
$\Rightarrow P\left( k \right)=\dfrac{11}{32}$
For k = 3, the probable outcome will be,
\[\Rightarrow E\left( k \right)=\left\{ HHHTT,THHHT,TTHHH,HHHTH,HTHHH \right\}=5\]
$\Rightarrow P\left( k \right)=\dfrac{5}{32}$
For k = 4, the probable outcome will be,
\[\Rightarrow E\left( k \right)=\left\{ HHHHT,THHHH \right\}=2\]
$\Rightarrow P\left( k \right)=\dfrac{2}{32}$
For k = 5, the probable outcome will be,
\[\Rightarrow E\left( k \right)=\left\{ HHHHH \right\}=1\]
$P\left( k \right)=\dfrac{1}{32}$
For k = 0, 1, 2, X = -1 and for k = 3, 4, 5, X = k.
The formula for the expected value of X is,
$\sum{XP\left( k \right)}$
Substitute the values in the above equation,
\[\Rightarrow \sum{XP\left( k \right)}=\left( -1 \right)\times \dfrac{1}{32}+\left( -1 \right)\times \dfrac{12}{32}+\left( -1 \right)\times \dfrac{11}{32}+3\times \dfrac{5}{32}+4\times \dfrac{2}{32}+5\times \dfrac{1}{32}\]
Multiply the terms,
\[\Rightarrow \sum{XP\left(k \right)}=-\dfrac{1}{32}-\dfrac{12}{32}-\dfrac{11}{32}+\dfrac{15}{32}+\dfrac{8}{32}+\dfrac{5}{32}\]
Since the denominator is same. Add and subtract the terms in numerator,
\[\Rightarrow \sum{XP\left( k \right)}=\dfrac{4}{32}\]
Cancel out the common factors from the numerator and denominator,
\[\Rightarrow \sum{XP\left( k \right)}=\dfrac{1}{8}\]
Thus, the expected value of X is $\dfrac{1}{8}$.

Hence, option (A) is correct.

Note:
The students might make mistake in calculating the occurrence of heads simultaneously. As the occurrence of heads and occurrence of heads simultaneously is a different thing.
A probability of 0 means that an event is impossible.
A probability of 1 means that an event is certain.
An event with a higher probability is more likely to occur.
Probabilities are always between 0 and 1.