# probability of three units

by pride welile
(kazan state,Russia)

7 regular cast of the dice. Find the probability that falls three units, two triples and at least one six.

### Comments for probability of three units

 Apr 29, 2011 Probability – Rolling Dice by: Staff ---------------------------------------------------- Part III ---------------------------------------------- Special Note: The theoretical probability and the experimental probability (also called the empirical probability, or observed probability) are not the same. Be aware that if you computed the experimental probability by actually rolling the dice yourself a large number of times (say, 700 times), then compared your computations with the theoretical probability, there would be a difference. At that point, you might be tempted to say something like this to yourself: “I didn’t roll the dice enough times. If I rolled the dice enough times, the theoretical probability and the experimental probability would be the same. That is not true. The theoretical probability and the experimental probability will never be the same no matter how many times the dice are rolled. If you performed a larger experiment by actually rolling the dice (say, 70,000 times, or 4,900,000,000 times), the experimental probability you compute WOULD NOT converge on the theoretical probability. Your experiments will converge on different number because of the weight distribution within the dice. Your experiments will be consistent with one another, but they will not be equal with the theoretical probability. The real (experimental) probability would be very important to you if you made your living gambling (which I hope you don’t). ---------------------------------------------- Thanks for writing. Staff www.solving-math-problems.com

 Apr 29, 2011 Probability – Rolling Dice by: Staff ---------------------------------------------------- Part II Breaking these sets down into the groups in your statement of the problem: {same digit1, same digit1, same digit1, same digit2, same digit2, same digit2, 6} (see list below): there are 8 sets {1, 1, 1, 2, 2, 2, 6} {1, 1, 1, 3, 3, 3, 6} {1, 1, 1, 4, 4, 4, 6} {1, 1, 1, 5, 5, 5, 6} {2, 2, 2, 3, 3, 3, 6} {2, 2, 2, 5, 5, 5, 6} {3, 3, 3, 4, 4, 4, 6} {3, 3, 3, 5, 5, 5, 6} {same digit1, same digit1, same digit1, 6, 6, 6, any other digit} (see list below): there are 16 sets {1, 1, 1, 2, 6, 6, 6} {1, 1, 1, 3, 6, 6, 6} {1, 1, 1, 4, 6, 6, 6} {1, 1, 1, 5, 6, 6, 6} {1, 3, 3, 3, 6, 6, 6} {1, 4, 4, 4, 6, 6, 6} {1, 5, 5, 5, 6, 6, 6} {2, 2, 2, 5, 6, 6, 6} {2, 3, 3, 3, 6, 6, 6} {2, 5, 5, 5, 6, 6, 6} {3, 3, 3, 4, 6, 6, 6} {3, 3, 3, 5, 6, 6, 6} {3, 4, 4, 4, 6, 6, 6} {3, 5, 5, 5, 6, 6, 6} {4, 4, 4, 5, 6, 6, 6} {4, 5, 5, 5, 6, 6, 6} Each of these sets can be arranged in different in different ways {same digit1, same digit1, same digit1, same digit2, same digit2, same digit2, 6} Calculate the number of ways this set can be arranged: Ways the digits in this number set can be arranged = 7!, if there were no duplicates Adjust for duplicates = 7!/(3!)(3!) = 7*6*5*4*3*2*1/[(3*2*1)(3*2*1)] = 7*6*5*4/(3*2) = 7*6*5*4/6 = 7*5*4 = 140 different arrangements of the 7 digits Probability = 140/279936 = .000500114 = .05% for EACH of the 8 sets in this category {same digit1, same digit1, same digit1, 6, 6, 6, any other digit} Ways the digits in this number set can be arranged = 7!, if there were no duplicates Adjust for duplicates = 7!/(3!)(3!) = 7*6*5*4*3*2*1/[(3*2*1)(3*2*1)] = 7*6*5*4/(3*2) = 7*6*5*4/6 = 7*5*4 = 140 different arrangements of the 7 digits Probability = 140/279936 = .000500114 = .05% for EACH of the 16 sets in this category The total probability of all the outcome sets is: P = 24 sets * .05% = 1.2 % The final answer is: 1.2% ----------------------------------------------------

 Apr 29, 2011 Probability – Rolling Dice by: Staff Part I The question: by Pride Welile (Kazan State, Russia) 7 regular cast of the dice. Find the probability that falls three units, two triples and at least one six. The answer: I’m going to compute the theoretical probability for your question. Find the probability that falls three units, two triples and at least one six. I’m not sure I understand what you are asking. I think you are asking what the probability is for the following sets in the outcome: If you throw 1 dice (or die, if you prefer) 7 times {same digit1, same digit1, same digit1, same digit2, same digit2, same digit2, 6} or {same digit1, same digit1, same digit1, 6, 6, 6, any other digit} When 7 dice are rolled ( or 1 dice 7 times), there are 6^n possibilities in the sample space. Since n=7, 6^7 = 279936 possibilities If order does not matter, 24 sets meet your criteria: {1, 1, 1, 2, 2, 2, 6} {1, 1, 1, 2, 6, 6, 6} {1, 1, 1, 3, 3, 3, 6} {1, 1, 1, 3, 6, 6, 6} {1, 1, 1, 4, 4, 4, 6} {1, 1, 1, 4, 6, 6, 6} {1, 1, 1, 5, 5, 5, 6} {1, 1, 1, 5, 6, 6, 6} {1, 3, 3, 3, 6, 6, 6} {1, 4, 4, 4, 6, 6, 6} {1, 5, 5, 5, 6, 6, 6} {2, 2, 2, 3, 3, 3, 6} {2, 2, 2, 5, 5, 5, 6} {2, 2, 2, 5, 6, 6, 6} {2, 3, 3, 3, 6, 6, 6} {2, 5, 5, 5, 6, 6, 6} {3, 3, 3, 4, 4, 4, 6} {3, 3, 3, 4, 6, 6, 6} {3, 3, 3, 5, 5, 5, 6} {3, 3, 3, 5, 6, 6, 6} {3, 4, 4, 4, 6, 6, 6} {3, 5, 5, 5, 6, 6, 6} {4, 4, 4, 5, 6, 6, 6} {4, 5, 5, 5, 6, 6, 6} ----------------------------------------------------