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) ⇒0−−−−−−−⇒ each blank can be filled in 2 ways either 1 or 0 then, P(0)=27=128 combinations is possible case 2: Probability of number having last bits as 11 can be given as P(11) ⇒−−−−−−11⇒ each blank can be filled in 2 ways either 1 or 0 then, P(11)=26=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) ⇒0−−−−−11⇒ each blank can be filled in 2 ways either 1 or 0 then, P(0&11)=25=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