If a student can solve one of 4 problems. If he attempts 4 questions in succession. What is the probability that at least one of them is right?

• p(s)=1/4
  p(s')=3/4
  prob of not solvng ny problem is 3/4*3/4*3/4*3/4=81/256
  so prob tht he slv 1ques is=1-81/256=175/256
• 10 years agoHelpfull: Yes(45) No(0)
• Prob. that he can solve at least one = 1 - prob(solving none)^n
  
  Prob(solving 1 problem out of 4) = 1/4
  Prob(solving none) = 1 - 1/4 => 3/4
  Therefore Prob(atleast one) = 1 - (3/4)^4
  => 1 - 81/256
  => 175/256
  
  Ans : 175/256
• 10 years agoHelpfull: Yes(11) No(0)
• p(s)=1/4
  p(s')=3/4
  Prob that atleast one of them is right= (1/4*3/4*3/4*3/4)+(1/4*1/4*3/4*3/4)+(1/4*1/4*1/4*3/4)+(1/4*1/4*1/4*1/4)
  = 27/256+ 9/256+ 3/256+ 1/256
  =40/256
• 10 years agoHelpfull: Yes(8) No(4)
• p(s)=1/4
  p(s')=3/4
  student can solve only one problem
  so the cases are
  when 1st problem is solved or 2nd is solver or 3rd is solved or 4th is solved
  = (1/4*3/4*3/4*3/4)+(3/4*1/4*3/4*3/4)+(3/4*3/4*1/4*3/4)+(3/4*3/4*3/4*1/4)
  =27/64
• 10 years agoHelpfull: Yes(5) No(4)
• p(s)=1/4
  p(~s)=3/4
  prob of not solvng ny problem is 3/4*3/4*3/4*3/4=81/256
  so prob tht he slv 1ques is=1-81/256=175/256
• 10 years agoHelpfull: Yes(2) No(0)
• when one of the 4 is right
  p=1/4*3/4*3/4*3/4=27/256
  when 2 of them is right
  p=1/4*1/4*3/4*3/4=9/256
  when 3 of 4 is right
  p=1/4*1/4*1/4*3/4=3/256
  when all are right
  p=1/256
  so
  p(S) = 1/256 +3/256 +9/256 + 27/256 =40/256 =5/32
• 10 years agoHelpfull: Yes(2) No(1)
• It may be 1.. I'm nt sure.
• 10 years agoHelpfull: Yes(1) No(8)
• ->>> way 1
  ---------
  
  p(solving at least one of the problem right) = 1- p(solving none of the problem right)
  p(solving) = 1/4
  p(not solving)=1-1/4=3/4
  p(solving none of the problem right)= p(all 4 problems wrong)
  =p(not solving)^4
  =(3/4)^4
  =81/256
  
  p(solving at least one of the problem right) = 1 - 81/256
  = 175/256
  ---->> way 2
  -------------------
  p(s)=1/4
  p(s')=3/4
  p(solving at least one of the problem right) = p(1 right problem) + p(2 right problem)
  +p(3 right problem) + p(4 right problem)
  p(1 right problem) = 4C1 * p(s) * 3C3 * p(s')^3 = 4 * (1/4) * (3/4)^3 = 108/356
  p(2 right problem) = 4C2 * p(s)^2 * 2C2 * p(s')^2 = 6 * (1/4)^2 * (3/4)^2 = 54/256
  p(3 right problem) = 4C3 * p(s)^3 * 1C1 * p(s') = 4 * (1/4)^3 * (3/4) = 12/256
  p(4 right problem) = 4C4 * p(s) ^4 = 1/256
  p(solving at least one of the problem right) = (108 + 54 +12 + 1)/256
  = 175/256
• 9 years agoHelpfull: Yes(1) No(0)
• 1-(3/4)^4=135/216
• 10 years agoHelpfull: Yes(0) No(1)
• Every one has his/her own answer,what kind of this solutin is...
  
• 10 years agoHelpfull: Yes(0) No(0)
• he attempts all 4 que
  here it is at least 1 so right ques can be 1 or 2 or 3 or 4
  so prp of sucess=1/4*2/4*3/4*4/4=3/32
• 10 years agoHelpfull: Yes(0) No(2)