The answer is 1/12 because tbelow are 2 methods of rolling 11, rolling a 5 and also a 6 or rolling a 6 and a 5 (for probcapability objectives there are indeed 2 different results). There is only 1 way to roll over 11 (2 6's). That suggests that out of 36 feasible combicountries, 3 qualify as being equal to or better than 11, which implies that the probcapacity is 3/36, which reduces to 1/12.
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The answer is tright here are 6 sides on each die, therefore, 6 times 6 is 36, giving us all the feasible outcomes. You deserve to roll an 11 among 2 means a, 5 and also a 6 or a 6 and also a 5. You have the right to roll a 12 just one means. This offers us 3 possible outcomes of 11 or over out of 36. Which reduces to 1/12.
A single die has 6 possible sides it deserve to land also on. That suggests 2 die have actually 36 feasible outcomes, yet just 18 distinctive. The probability that the 2 rolls die amount to 11 is 1/18He answer is 1/18, not 1/12.
*the answer is 1/18 for the amount of an 11 and also 1/9 for 11 and over.The correct answer is 1/9.(5,6)-(6,5)-(6,6)-(6,6) 4/36=1/9
Casey's appropriate, 3/36. Peter, it isn't an ordered problem yet tright here are 2 rolls out of the 36 feasible that deserve to be 11 (6 on die a, 5 on die b, and also vice versa). Add that to the one method you have the right to roll boxcars (6 on die a, 6 on b), entirely of 3 winning instances out of 36 feasible.
OK, you are right Casey...I made myself think that what matters is the final sum, so unmuch less they come up with dice that have 12 sides...I'm wrong :)Thanks Grant
"Dices"? Did you just say "dices"? Excuse me for answering your question via a question, however what was the greatest grade of grammar school education and learning you recontact passing? If I gained this task, is there any possibility, that is, any opportunity higher than zero, that you would certainly be my boss? How did you get to your present place in this organization? Can I go now?
I'm via Jeff and also Carlos- "dice" is the plural of "die". Geez. However before, if the question actually is "what is the probcapacity of obtaining 11 or higher on a solitary throw via a pair of dice," then the answer is 50%. It either will or will not be 11 or higher.
1) Same as Jeff - "dices" provides no sense2) How many sides are tbelow on each die? Don't assume that all dice have actually 6 sides.
All possible results as soon as throwing 2 dices ( 2-3 combination IS THE SAME as 3-2, you understand it if you ever played dice, I don't know what are the probability objectives Casey is talking around ):(1,1)(1,2) (2,2)(1,3) (2,3) (3,3)(1,4) (2,4) (3,4) (4,4)(1,5) (2,5) (3,5) (4,5) (5,5)(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)so there are 21 combinations, only (5,6) and (6,6) are >=11,so the answer is: 2/21=9.52%
I think the question is oriented to find out if you are actually listening. You deserve to not throw 11 AND over. You can throw 11 OR over. So, the answer is 0. Impossible.
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If you are rolling 2 dice the chances of obtaining an 11 and OVER is 2/12= 1/6. You deserve to roll a 6 and a 5 to equal 11 and you can roll a 6 and also a 6 and get 12 which is OVER 11, therefore the answer have to be 1/6.
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