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 $P(0)$ $\Rightarrow 0_{\text{-------}} \Rightarrow$ each blank can be filled in 2 ways either 1 or 0 then, $P(0)=2^{7}=128$ combinations is possible case 2: Probability of number having last bits as 11 can be given as $P(11)$ $\Rightarrow_{\text{------}} 11 \Rightarrow$ each blank can be filled in 2 ways either 1 or 0then, $P(11)=2^{6}=64$ combinations is possible case 3: Probability of number having first bit as 0 last bits as 11 can be given as $P(0 \& 11)$ $\Rightarrow 0_{\text{-----}} 11 \Rightarrow$ each blank can be filled in 2 ways either 1 or 0 then, $P(0 \& 11)=2^{5}=32$ 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 $P(T)$ $P(T)=P(0)+P(11)-P(0 \& 11)$ $P(T)=128+64-32=160$