A and B are 2 independent events . the probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. the probability of occurence of A is

• Independent:P(A^B)=P(A).P(B)=1/6 (^ stands for intersection)
  neither occurs:P(A' ^ B')=P(A u B)'=1/3 (De-Morgan's Law),P(A u B)=1-1/3=2/3
  P(A u B)=P(A)+P(B)-P(A)P(B)=>2/3=P(A)+1/6(P(A))-1/6,solving P(A)= 1/3 or 1/2
• 10 years agoHelpfull: Yes(24) No(1)
• probability of occuring atlist one of them will be 1-1/3=2/3
  now probability of occurung only a or only b is=2/3 -1/6=1/2
• 10 years agoHelpfull: Yes(14) No(0)
• A+B=1/6
  A-B=1/3
  SO A=1/4
• 10 years agoHelpfull: Yes(3) No(3)
• is it 1/3 or 1/2
  
• 10 years agoHelpfull: Yes(1) No(0)
• Given:
  P(A and B)=1/6=P(A).P(B)(independent events)
  P(A' and B')=1/3
  Now , P(A' and B')=(1-P(A))(1-P(B))= 1-P(A)-P(B)+P(A)P(B)=1/3
  =>P(A)+P(B)=5/6----(I)
  1/6=P(A).P(B)----(ii)
  SOLVING ABOVE TWO EQUATIONS FOR P(A),
  P(A)+1/(6(P(A))) =5/6
  =>P(A)=1/2 OR 1/3
• 10 years agoHelpfull: Yes(1) No(0)