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What is the probability of rolling a total of 7 with a singl
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15 Oct 2018, 14:11
15 Oct 2018, 14:11
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What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1– 6 on each side?
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
Re: What is the probability of rolling a total of 7 with a singl
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15 Oct 2018, 20:12
15 Oct 2018, 20:12
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Carcass wrote:
What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1– 6 on each side?
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
For rolling a total of 7 with a single roll of two fair six-sided dice, there are 6 possibilities
First dice & second dice:
6 + 1 , 1 + 6
2 + 5 , 5 + 2
3 + 4 , 4 + 3
Total possible outcomes =\(6 * 6 = 36\)
Hence the probability of rolling a total of 7 with a fair six sided dice = \(\frac{6}{36}=\frac{1}{6}\)
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Re: What is the probability of rolling a total of 7 with a singl
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21 Jun 2019, 08:18
21 Jun 2019, 08:18
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Carcass wrote:
What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1– 6 on each side?
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
A. \(\frac{1}{12}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{7}\)
D. \(\frac{1}{3}\)
E. \(\frac{1}{2}\)
Here's a solution that uses probability rules (rather than counting methods)
Key concept: when it comes to getting a sum of 7, we can roll ANY number on the first roll. That is, no matter what number is rolled 1st, that number can be paired with another die number to get a sum of 7. What matters is the 2nd role.
Given this:
P(sum is 7) = P(1st roll is ANY number AND 2nd roll is the one number that creates a sum of 7 with the 1st number)
= P(1st roll is ANY number) x P(2nd roll is the one number that creates a sum of 7 with the 1st number)
= 1 x 1/6
= 1/6
= B
Cheers,
Brent
