What is the probability of getting at least one six in a single throw of three unbiased dice?

• for no 6 = 5*5*5 / 6*6*6 = 125/216
  
  for atleast one 6 = 1-125/216 = 91/216.
• 12 years agoHelpfull: Yes(12) No(2)
• Find the number of cases in which none of the digits show a '6'.
  
  i.e. all three dice show a number other than '6', 5 * 5 *5 = 125 cases.
  
  Total possible outcomes when three dice are thrown = 216.
  
  The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
  
  = 216 - 125 = 91.
  
  The required probability = 91/216
• 13 years agoHelpfull: Yes(9) No(0)
• P(getting a 6) = 1/6
  P(not getting a 6 on one throw) = 5/6
  P (not getting a 6 on three throws) = 5/6*5/6*5/6 = 125/216
  
  P(getting at least a 6) = 1-125/216= 91/216
• 9 years agoHelpfull: Yes(4) No(0)
• P(getting a 6) = 1/6
  P(not getting a 6 on one throw) = 5/6
  P (not getting a 6 on three throws) = 5/6*5/6*5/6 = 125/216
  
  P(getting at least a 6)=1-125/216=91/216
• 9 years agoHelpfull: Yes(0) No(0)