Persons A and B. Person A picks a random no. from 1 to 1000.Then person B picks a random no. from 1 to 1000. What is the probability of B getting no. greater then what A has picked.

• Probability of A choosing 1 and B greater = (1/1000)*(999/1000);
  Probability of A choosing 2 and B greater = (1/1000)*(998/1000);
  Probability of A choosing 3 and B greater = (1/1000)*(997/1000); ...;
  Probability of A choosing 998 and B greater = (1/1000)*(2/1000);
  Probability of A choosing 999 and B greater = (1/1000)*(1/1000);
  Total = (1/1000)*(1/1000)*(999+998+997+...+2+1) = (1/1000)*(1/1000)*(999*1000/2) = 0.4995
• 14 years agoHelpfull: Yes(15) No(0)
• EASIEST WAY S SOLVING USING FORMULA ;
  prob of B getting greaterr no than A= P(B)/ ( p(A)+p(B))
  P(A)=1/1000 P(B)=1/1000
  Therefore ans = (1/1000) / (2/1000) = 1/2
• 8 years agoHelpfull: Yes(7) No(0)
• Pbbl of A choosing 1 = (1/1000). Pbbl of B choosing >1 = (999/1000). Total = (1/1000)*(999/1000) and so on. Total pbbl = (1/1000)*(999/1000)+(1/1000)*(998/1000)+...+(1/1000)*(1/1000) = (1/1000)*(1/1000)*(999+998+...+2+1) = 0.4995
• 14 years agoHelpfull: Yes(4) No(0)
• 999/2000
• 14 years agoHelpfull: Yes(3) No(1)
• (1/1000)*(999/999) + (1/1000)*(998/999) + ... + (1/1000)*(1/999) = (1/1000)*(1/999)*[999+998+...+2+1] = (1/1000)*(1/999)*[(999*1000)/2] = (1/2)
• 14 years agoHelpfull: Yes(2) No(0)