if 3 dice are tossed then what is the probability of getting atleast one six ?>
plz tell me answer ???????????

• 1-((5/6)*(5/6)*(5/6))
• 10 years agoHelpfull: Yes(37) No(3)
• six other other + six six other + six six six
  (1/6)(5/6)(5/6)*3!/2! + (1/6)(1/6)(5/6)*3!/2! + (1/6)(1/6)(1/6)*3!/3!
  =91/216
• 10 years agoHelpfull: Yes(28) No(2)
• Atleast 1 six means: it may be one six or two six or all three six :
  
  when exactly one dice(D1) occupy 1 six, then other should not equals to six so possibility=(5*5=25)
  (6 1 1)...(6 1 5) = total 5 set
  (6 2 1)...(6 2 5) = total 5 set
  (6 3 1)...(6 3 5) = total 5 set
  (6 4 1)...(6 4 5) = total 5 set
  (6 5 1)...(6 5 5) = total 5 set, so clear total 25 set with only one six.its all for 1st dice(D1) here 2nd dice(D2) or 3rd dice(D3) may hold one six then total possibility =25+25+25 =75
  
  now lets consider two six on (D1 D2)
  (6 6 1)..(6 6 5)= 5 set. It may happen for (D1 D3) or (D2 D3) so tatal probability]= 5+5+5=15
  
  lets consider all three dice hold six, possibility =1
  
  so total evevnt = 75+15+1=91
  bcz total sample =6*6*6=216
  
  Probability = 91/216
  so total 25+5+1=31
• 10 years agoHelpfull: Yes(9) No(2)
• possible favorable outcomes for at least one six are 7
  0 0 6
  0 6 0
  6 0 0
  0 6 6
  6 0 6
  6 6 0
  6 6 6
  probability = 7/216
• 10 years agoHelpfull: Yes(7) No(25)
• Three cases arise:
  
  Case 1: When only one dice shows up a six
  
  This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability
  
  Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)
  
  =(1/6)*(5/6)*(5/6)
  
  =25/216
  
  similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216
  
  And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216
  
  So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216
  
  Case 2:When two dice show up a six
  
  Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3
  
  Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs
  
  Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216
  
  So, total probability = 3*(5/216) = 15/216
  
  Case 3: When all dice show up a six
  
  In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216
  
  
  
  So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• Three cases arise:
  
  Case 1: When only one dice shows up a six
  
  This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability
  
  Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)
  
  =(1/6)*(5/6)*(5/6)
  
  =25/216
  
  similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216
  
  And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216
  
  So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216
  
  Case 2:When two dice show up a six
  
  Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3
  
  Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs
  
  Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216
  
  So, total probability = 3*(5/216) = 15/216
  
  Case 3: When all dice show up a six
  
  In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216
  
  
  
  So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216
• 10 years agoHelpfull: Yes(3) No(0)
• d) 91/216
  
• 10 years agoHelpfull: Yes(2) No(3)
• (3C1+3C2+3C3) /6*6*6==7/216
• 10 years agoHelpfull: Yes(1) No(1)
• In the first throw there can be six or other no. so, for atleast condition
  1/6*5/6*5/6*3(first throw result in a six and other does not and this can be arranged in 3 ways)+1/6*1/6*5/6*3(case in which two throw result in a six and their possible arrangement)+1/6*1/6*1/6(case in which all throw result in a six)
• 10 years agoHelpfull: Yes(0) No(0)
• why its not 31/216 can any1 explain me plzzz
• 10 years agoHelpfull: Yes(0) No(0)