Three Fair DiceDate: 05/16/99 at 08:01:51 From: Laura Subject: Probability dealing with combinations of dice Dear Dr. Math, I have been trying to solve this problem for quite a while but I have not been able to come up with an answer. a) Find the probability of a total of 9 and the probability of a total of 10 when three fair dice are tossed once. Compare the probabilities. b) Find the probability of at least one six in 4 tosses of a fair die and the probability of at least one double-six in 24 tosses of two fair dice. Compare the probabilities. c) Find the probability of at least one six when 6 dice are rolled and the probability of at least two sixes when 12 dice are rolled. Compare the probabilities. For a), I found that there are 342 possible total outcomes when rolling 3 dice, but I do not think this is correct. I got 342 by writing out all the possible outcomes - I counted the triple digits (i.e. getting a 3, a 3, and a 3) as six ways instead of one. Also, is it correct to take the number of ways to get a nine and divide by the total possible outcomes, to find the probability of getting a nine? For b), I again was not sure how to solve this. Do I take 1/6, the probability of getting a 6, and multiply it by 4 - the number of times tossed? And for the second part, I have no idea. For c), is it 1/6 times 6 or 1/6 ^6, or.....? Finally, how do I compare? Would it be reasonable to take the two outcomes and state their difference? What, do you think, should I include in the statement? If you could answer even part of this question, I would greatly appreciate it. Thank you very much. Laura Elisibeth Mazzeo 11th grade IB student Date: 05/16/99 at 16:24:04 From: Doctor Anthony Subject: Re: Probability dealing with combinations of dice a) The generating function is (x + x^2 + x^3 + x^4 + x^5 + x^6)^3 and you require the coefficient of x^9 and x^10 The term in brackets is [x(1-x^6)]^3 ------------- = x^3(1 - 3x^6 + 3x^12 - x^18)(1-x)^(-3) (1-x)^3 = (x^3 - 3x^9 + 3x^15 - x^21)(1 + C(3,1)x + C(4,2)x^2 + C(5,3)x^3 +.. We must pick out the terms in x^9 x^3.C(8,6)x^6 - 3x^9 Required coefficient = C(8,6) - 3 = 28 - 3 = 25 And the coefficient of x^10 x^3.C(9,7)x^7 - 3x^9.C(3,1)x Required coefficient = C(9,7) - 9 = 36 - 9 = 27 The number of outcomes with 3 dice is 6^3 = 216, so we get Probability of a total of 9 = 25/216 Probability of a total of 10 = 27/216 = 1/8 b) In the first situation we need the probability of AT LEAST one six in 4 throws. That is 1 - P(0) where P(0) = prob. no six in 4 throws P(0) = (5/6)^4 = 0.48225 So the probability of at least 1 six = 0.51775 For the second situation the probability of a double six is 1/36 Probability of no double six = 35/36 In 24 throws P(0) = (35/36)^24 = 0.508596 and therefore the probability of at least one double six = 0.4914 The probabilities are 0.51775 for at least one six in 4 throws and a probability of 0.4914 for at least one double six in 24 throws. c) Probability of no six with 6 dice = (5/6)^6 = 0.334898 So the probability of at least one six = 0.6651 Probability of no six with 12 dice = (5/6)^12 = 0.1121566 Probability of 1 six with 12 dice = C(12,1)(1/6)(5/6)^11 = 0.26917597 So the probability of 0 or 1 six with 12 dice = 0.38133257 Probability of at least 2 sixes with 12 dice = 0.618667 Therefore the probability is 0.6651 for at least one six with 6 dice and the probability is 0.618667 for at least 2 sixes with 12 dice. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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