NIMCET 2014 Question Paper with solutions

Consider the 8 -bit number with 8 blanks as shown --------
case 1: Probability of number having first bit as 0 can be given as $\mathit{P}\left(0\right)$
$\Rightarrow 0_{-------}\Rightarrow$ each blank can be filled in 2 ways either 1 or 0
then, $\mathit{P}\left(0\right)=2^7=128$ combinations is possible
case 2: Probability of number having last bits as 11 can be given as $\mathit{P}\left(11\right)$
${\Rightarrow }_{------}11\Rightarrow$ each blank can be filled in 2 ways either 1 or 0
then, $\mathit{P}\left(11\right)=2^6=64$ combinations is possible
case 3: Probability of number having first bit as 0 last bits as 11 can be given as
$\Rightarrow 0_{-----}11\Rightarrow$ each blank can be filled in 2 ways either 1 or 0
then, combinations is possible
Finally, the total number of combinations possible for having an 8 -bit number that starts with the bit 0 or ends with the bits 11 is $\mathit{P}\left(\mathit{T}\right)$

$\mathit{P}\left(\mathit{T}\right)=128+64-32=160$