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**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.

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