A student is given a test with \[8\] items of true-false type. If he gets \[6\] or more items correct, he is declared a pass. Given that he guesses the answer to each item, compute the probability that he will pass in the test.

Hint:
Here we need to apply the concept of Binomial Theorem involved with probability. Exponent rules and factorial definition is needed. OR rule in probability is to be applied for adding the events.
Formula Required: $Probability\left( event \right)=\dfrac{Favourable\text{ }cases}{Total\text{ }cases}$
Binomial theorem states $P(x)={{n}_{{{C}_{x}}}}\times {{p}^{x}}\times {{\left( 1-p \right)}^{n-x}}$
$P(x)$means Total required probability
$x=$ Total number of failure cases.
\[n\text{ }=\] Total number of trials.
$n-x$ means number of failure cases.
\[p\] means probability of success of an event.
\[1-p\] means probability of failure of an event.
$\begin{align}
  & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\
 & {{a}^{0}}=1 \\
\end{align}$
${{n}_{{{C}_{r}}}}=\dfrac{n!}{r!\left( n-r \right)!}$ , r selections out of n.
$\begin{align}
  & n!=n\times n-1\times n-2\times ..........\times 3\times 2\times 1 \\
 & 0!=1 \\
\end{align}$
We need to find the probability that he will pass the test.

Complete step by step solution:
Probability of answering true for each item with either true or false is,
$\begin{align}
  & P\left( True \right)=\dfrac{Number\text{ }of\text{ }option\operatorname{s}\text{ }in\text{ }favour\text{ }as\text{ }true}{Total\text{ }Number\text{ }of\text{ }option\operatorname{s}} \\
 & =\dfrac{1}{2} \\
 & P\left( False \right)=\dfrac{Number\text{ }of\text{ }option\operatorname{s}\text{ }in\text{ }favour\text{ }as\text{ }false}{Total\text{ }Number\text{ }of\text{ }option\operatorname{s}} \\
 & =\dfrac{1}{2} \\
\end{align}$
According to the question,
Pass means answering $6$ or more questions as true
Required Probability is answering $6$ or more questions as true,
$\Rightarrow $ $\begin{align}
  & P\left( Pass \right)=P\left( 6T,2F \right)orP\left( 7T,1F \right)orP\left( 8T,0F \right) \\
 & P\left( Pass \right)=P\left( 6T,2F \right)+P\left( 7T,1F \right)+P\left( 8T,0F \right) \\
\end{align}$
Where T and F are questions answered as True and false respectively.
Applying Binomial Theorem,
\[\begin{align}
  & P\left( 6T,2F \right)={{8}_{{{C}_{6}}}}\times {{\left( P\left( T \right) \right)}^{6}}\times {{\left( P\left( F \right) \right)}^{2}} \\
 & =\dfrac{8!}{6!\left( 8-6 \right)!}\times {{\left( \dfrac{1}{2} \right)}^{6}}\times {{\left( \dfrac{1}{2} \right)}^{2}} \\
 & =\dfrac{8\times 7\times 6!}{6!\left( 2 \right)!}\times {{\left( \dfrac{1}{2} \right)}^{8}} \\
 & =28\times \dfrac{1}{256} \\
 & =\dfrac{28}{256} \\
\end{align}\]

\[\begin{align}
  & P\left( 7T,1F \right)={{8}_{{{C}_{7}}}}\times {{\left( P\left( T \right) \right)}^{7}}\times {{\left( P\left( F \right) \right)}^{1}} \\
 & =\dfrac{8!}{7!\left( 8-7 \right)!}\times {{\left( \dfrac{1}{2} \right)}^{7}}\times {{\left( \dfrac{1}{2} \right)}^{1}} \\
 & =\dfrac{8\times 7!}{7!}\times {{\left( \dfrac{1}{2} \right)}^{8}} \\
 & =8\times \dfrac{1}{256} \\
 & =\dfrac{8}{256} \\
\end{align}\]
\[\begin{align}
  & P\left( 8T,0F \right)={{8}_{{{C}_{8}}}}\times {{\left( P\left( T \right) \right)}^{8}}\times {{\left( P\left( F \right) \right)}^{0}} \\
 & =\dfrac{8!}{8!\left( 8-8 \right)!}\times {{\left( \dfrac{1}{2} \right)}^{8}}\times {{\left( \dfrac{1}{2} \right)}^{0}} \\
 & =\dfrac{8!}{8!}\times {{\left( \dfrac{1}{2} \right)}^{8}} \\
 & =1\times \dfrac{1}{256} \\
 & =\dfrac{1}{256} \\
\end{align}\]
Required Probability ,
$\begin{align}
  & P\left( Pass \right)=P\left( 6T,2F \right)+P\left( 7T,1F \right)+P\left( 8T,0F \right) \\
 & =\dfrac{28}{256}+\dfrac{8}{256}+\dfrac{1}{256} \\
 & =\dfrac{37}{256} \\
\end{align}$
Therefore, the probability that he will pass in the test is $\dfrac{37}{256}$ .

Note:
In such types of questions the concept of concept Binomial Theorem involved with probability . Basic formula of Probability is needed. While applying Binomial Theorem formula is needed. Appropriate data is applied in the formula to get the required answer.