What is the probability of rolling a total of 7 with two dice at least once in 10 rolls?

Question

Example: A 5 and a 2 on the dice equal a total of 7.

Answer 1

First find the probability of rolling a seven with one roll of two dice.

P(7) = (# of ways to roll 7) / (number of ways to roll 2 dice)
P(7) = 6/36 = 1/6

P(no 7 in one roll) = 1 - 1/6 = 5/6

The probability of at least one seven in ten rolls of two dice is one minus the probability of no sevens.

P(at least one 7) = 1 - (5/6)^10 ≈ .838

Answer 2

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Use this table to see all 36 outcomes of rolling two dice. The first column is the face showing on die1 and the first row is the face showing on die2. The other values in the table are the sum of the two dice. ___1___2___3___4___5___6 1__2___3___4___5___6___7 2__3___4___5___6___7___8 3__4___5___6___7___8___9 4__5___6___7___8___9___10 5__6___7___8___9___10__11 6__7___8___9___10__11__12 there is a 6/36 = 1/36 probability of rolling a 7, assuming the dice are fair. Let X be the number of 7's in 10 rolls of the die. Find P(X ≥ 1) = 1 - P(X = 0) P(X = 0) = P(no 7's rolled) = (1 - 1/6)^10 = (5/6)^10 = 9765625 / 60466176 = 0.1615056 P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.1615056 = 0.8384944 The more mathy away to do this and to generalize the probelm: X has the binomial distribution with n = 10 trials and success probability p = 1/6. In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures. Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials. X ~ Binomial( n , p ) the mean of the binomial distribution is n * p = 1.666667 the variance of the binomial distribution is n * p * (1 - p) = 1.388889 the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.178511 The Probability Mass Function, PDF, f(X) = P(X = x) is: P( X = 0 ) = 0.1615056 P( X = 1 ) = 0.3230112 P( X = 2 ) = 0.2907100 P( X = 3 ) = 0.1550454 P( X = 4 ) = 0.05426588 P( X = 5 ) = 0.01302381 P( X = 6 ) = 0.002170635 P( X = 7 ) = 0.0002480726 P( X = 8 ) = 1.860544e-05 P( X = 9 ) = 8.269086e-07 P( X = 10 ) = 1.653817e-08 The Cumulative Distribution Function, CDF, F(X) = P(X ≤ x) is: x ∑ P(X = t) = t = 0 P( X ≤ 0 ) = 0.1615056 P( X ≤ 1 ) = 0.4845167 P( X ≤ 2 ) = 0.7752268 P( X ≤ 3 ) = 0.9302722 P( X ≤ 4 ) = 0.984538 P( X ≤ 5 ) = 0.9975618 P( X ≤ 6 ) = 0.9997325 P( X ≤ 7 ) = 0.9999806 P( X ≤ 8 ) = 0.9999992 P( X ≤ 9 ) = 1 P( X ≤ 10 ) = 1 1 - F(X) is: n ∑ P(X = t) = t = x P( X ≥ 0 ) = 1 P( X ≥ 1 ) = 0.8384944 P( X ≥ 2 ) = 0.5154833 P( X ≥ 3 ) = 0.2247732 P( X ≥ 4 ) = 0.06972784 P( X ≥ 5 ) = 0.01546197 P( X ≥ 6 ) = 0.002438156 P( X ≥ 7 ) = 0.0002675215 P( X ≥ 8 ) = 1.944889e-05 P( X ≥ 9 ) = 8.434468e-07 P( X ≥ 10 ) = 1.653817e-08

Answer 3

Probability Of Rolling A 7

Answer 4

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RE:
What is the probability of rolling a total of 7 with two dice at least once in 10 rolls?
Example: A 5 and a 2 on the dice equal a total of 7.

Answer 5

The probability, of rolling at least one 7 in 10 rolls is one minus the probability that you never roll a 7. The probability of rolling a 7 on two 6-sided die in one roll is 1/6. So the probability of not rolling a 7 is 5/6. The probability of not rolling a 7 10 times in a row is (5/6)^10, or about 16.2%. So the probability of rolling at least one 7 is about 83.8%

Answer 6

There are 36 combinations in each roll and that means there 36^10 combinations
6 out of each 36 equal to 7 and that means we have 30 combinatins who don't equal 7

the answer is 1-(a/b)
where b=36^10
a=30^10

Answer 7

4 and 3

Answer 8

well probability of a seven is 6/36
1-6,6-1,5-2,2-5,3-4,4-3
so find the probability of not getting a 7 on 10 rolls
(5/6)^10=0.1615
so 1-0.1615=0.8385
so an 83.85%

Answer 9

My Anwser Is Not Likely