three dices are thrown what is d probabily to gat atleast one six?

• Probality of not getting any six is = (5/6)^3 = 125/216
  Therefore Probability of getting at least one six is = 1-125/216 = 91/216
• 12 years agoHelpfull: Yes(110) No(2)
• the probabilty of getting no six at all is (5/6)*(5/6)*(5/6)=125/216
  now probabilty of getting atleast 1 six is= 1-probabilty of getting no six at all
  = 1-125/216
  =91/216(Ans)
• 12 years agoHelpfull: Yes(39) No(0)
• Every die has got six sides. Each of the sides is numbered from 1 to 6.
  
  When a single unbiased die is thrown you can have six possible outcomes.
  
  When two dice are thrown simultaneously, the total number of outcomes will be 6 * 6 = 36
  Similarly, when three dice are thrown simultaneously, the total number of outcomes will be 6*6*6=216.
  
  We need to find out the number of cases in which at least one of the facing sides shows 6.
  
  At least one means - either one dice or two dice or all three.
  
  Case 1: Let us take the easiest case first - all three dice showing '6' - There is only one such possibility.
  
  Case 2: The number of cases in which two of the dice show 6 and one of them is a different number.
  For eg an event like 6 6 5 will be one of the outcomes for case 2.
  
  As two of the dice show '6' , it can happen in only one way. The third die shows a different number, a number other than 6, and it can be any of the 5 other numbers. Therefore, there will 5 possible options i.e. (1, 6, 6), (2, 6, 6), (3, 6, 6), (4, 6, 6), (5, 6, 6).
  
  However, each of these possibilities can have three different arrangements depending upon where the third different digit appears. For example take (1, 6, 6) case - it will have three options (1, 6, 6), (6, 1, 6), (6, 6, 1).
  
  Therefore, the total number of events in which 2 of the dice will show '6' and one will show a different number = 5*3 = 15 cases.
  
  Case 3: When only one of the die shows '6' and the other two show numbers other than '6'.
  
  The die showing '6' has only one option. The other two dice can have any of the '5' options. Therefore, the total number of possibilities = 1*5*5 = 25.
  
  However, the die showing '6' can either be the first die or the second die or the third die.
  
  Therefore, there are a total of 25 * 3 = 75 possibilities.
  
  Total possible outcomes where at least one die shows '6' = Case 1 + Case 2 + Case 3 = 1 + 15 +75 = 91.
  
  Therefore, the required probability = 91/216
• 10 years agoHelpfull: Yes(27) No(1)
• Here There Is Probability Of Getting 1,2,3 Sixs...So (1/6)*(5/6)(5/6)+(1/6)*(1/6)*(5/6)+(1/6)*(1/6)*(1/6)=31/216
• 12 years agoHelpfull: Yes(15) No(30)
• 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
  
  Answer: 91/216
• 9 years agoHelpfull: Yes(8) No(0)
• P(E)= number of possible events/number of sample space
  P(E)=7/216
• 12 years agoHelpfull: Yes(5) No(21)
• probablity to get any six when only one dice is thrown=1/6
  so,probablity of not getting any six=1-1/6=5/6
  now for 3 dice it becomes=(5/6)^3=125/216
  therefore probablity of getting atleast one six=1-125/216=91/216.
• 9 years agoHelpfull: Yes(4) No(0)
• favorable condition = 7
  total condition = 216
  probability = 7/216
• 10 years agoHelpfull: Yes(2) No(14)
• ans is. 91/216
• 9 years agoHelpfull: Yes(2) No(0)
• for no 6 number=5/6*5/6*5/6
  125/216
  atleast one 6 =1-125/216
  =91/216 answer
• 9 years agoHelpfull: Yes(2) No(0)
• How to prepare for elitmus exam from where i can prepare well for it please tell me
• 6 years agoHelpfull: Yes(1) No(0)
• Let, u hab to make a 3 digit num using nos. 1 to 6 (repetition allowed),
  Total = 6x6x6 =216
  Now, by only using nos. 1 to 5, we can make, 5x5x5= 125
  No. of 3 digit nums with atleast one 6 are, 216-125= 91,
  So, probability will be 91/216.
• 6 years agoHelpfull: Yes(1) No(0)
• ans is 36/216=1/6
  
  possible combinations are
  (611,612,613,614,615,616) ,(621,622........626),(631,632........636),(641,.........646),(651......656)(661...666)
  6+ 6+ 6+ 6+ 6+ 6
  
  =36
  
  alternate
  using p&c
  6 ( 6 * 6 ) = 36
  --- ----
  
  therefore, probability = (36/216) = (1/6)
• 9 years agoHelpfull: Yes(0) No(4)
• Correct ans: 91/216
• 7 years agoHelpfull: Yes(0) No(0)
• A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2*3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
  
  B. There are 5*3 combinations that you will get 2 6(six). Thus there is a 15/216=5/72 chance of getting a 2 6(six) when rolling 3 dice.
  
  C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice.
  
  D. There is a 75+15+1/216=91/216 chance of any of them happening.
• 7 years agoHelpfull: Yes(0) No(0)