In a digital circuit which was to implement (A B) + (A)XOR(B), the designer implements (A B) (A)XOR(B) What is the probability of error in it ?

• 
  A B AB A XOR B (AB)+(A XOR B) (AB) (A XOR B)
  0 0 0 0 0 0
  0 1 0 1 1 0
  1 0 0 1 1 0
  1 1 1 0 1 0
  
  so, we are getting 3 wrong answers out of 4
  
  probability of error is 3/4=75%
• 10 years agoHelpfull: Yes(5) No(0)
• The answer should be 3/4
  
  because
  
  A B C = A B D = A xor B C + D CD
  0 0 0 0 0 0
  0 1 0 1 1 0
  1 0 0 1 1 0
  1 1 1 0 1 0
  
  so total outcome is 4 and favourable case is 3 so,
  probability of Error shoud be 3/4 = 75%
• 12 years agoHelpfull: Yes(2) No(1)
• Let us take 2 input value A&B
  According to digital circuit implement (AB)+(A)XOR(B)
  but designer implement (AB).(A)XOR(B)
  So
  A B (AB)+(A)XOR(B) (AB).(A)XOR(B)
  0 0 0+0=0 0.0=0
  0 1 0+1=1 0.1=0
  1 0 0+1=1 0.1=0
  1 1 1+0=1 1.1=1
  
  The Favorable condition is last 3 output
  so probability should be 3/4.
  
• 9 years agoHelpfull: Yes(1) No(0)