What"s the probcapability of acquiring a total of \$7\$ or \$11\$ when a pair of fair dice is tossed?

I currently looked it up on the internet and my answer matched the same answer on a website. However, though I am confident that my solution is right, I am curious if there"s an approach in which I could compute this much faster considering that the photo listed below reflects exactly how time consuming that sort of technique would be. Thanks in advance.

You are watching: What is the probability of rolling a 7   For \$7\$, view that the initially roll doesn"t matter. Why? If we roll anything from \$1\$ to \$6\$, then the second roll have the right to always acquire a sum of \$7\$. The second dice has probability \$frac16\$ that it matches with the first roll.

Then, for \$11\$, I choose to think of it as the probcapability of rolling a \$3\$. It"s much less complicated. Why? Try inverting all the numbers in your die table you had actually in the image. Instead of \$1, 2, 3, 4, 5, 6\$, go \$6, 5, 4, 3, 2, 1\$. You need to see that \$11\$ and \$3\$ overlap. From below, simply calculate that tbelow are \$2\$ methods to roll a \$3\$: either \$1, 2\$ or \$2, 1\$. So it"s \$frac236 = frac118\$.

Key takeaways:

\$7\$ is constantly \$frac16\$ probabilityWhen asked to find probcapability of a larger number (choose \$11\$), discover the smaller sized equivalent (in this instance, \$3\$).
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answered Aug 14 "20 at 2:33 FruDeFruDe
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To calculate the possibility of rolling a \$7\$, roll the dice one at a time. Notice that it does not matter what the first roll is. Whatever before it is, there"s one possible roll of the second die that offers you a \$7\$. So the chance of rolling a \$7\$ has to be \$frac 16\$.

To calculate the chance of rolling an \$11\$, roll the dice one at a time. If the initially roll is \$4\$ or less, you have actually no possibility. The first roll will certainly be \$5\$ or more, keeping you in the ball game, through probcapacity \$frac 13\$. If you"re still in the round game, your opportunity of acquiring the second roll you need for an \$11\$ is aget \$frac 16\$, so the total chance that you will roll an \$11\$ is \$frac 13 cdot frac 16 = frac118\$.

Adding these two independent probabilities, the possibility of rolling either a \$7\$ or \$11\$ is \$frac 16+ frac118=frac 29\$.

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answered Aug 14 "20 at 2:21 Robert ShoreRobert Shore
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Gotta love stars and bars approach.

The number of positive integer solutions to \$a_1+a_2=7\$ is \$inom7-12-1=6\$. Therefore the probcapacity of acquiring \$7\$ from two dice is \$frac636=frac16\$.

For \$11\$ or any type of number better than \$7\$, we cannot continue exactly like this, given that \$1+10=11\$ is likewise a solution for instance, and we recognize that each roll cannot produce greater number than \$6\$. So we modify the equation a tiny to be \$7-a_1+7-a_2=11\$ where each \$a\$ is less than 7. This is equivalent to finding the variety of positive integers solution to \$a_1+a_2=3\$, which is \$inom3-12-1=2\$. Thus, the probcapability of gaining \$11\$ from two dice is \$frac236=frac118\$

Try to experiment through various numbers, calculate manually and also using various other approaches, then compare the result.

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answered Aug 14 "20 at 2:33 \$endgroup\$
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Welinvolved the jiyuushikan.org Stack Exchange.

Tbelow certain is a quicker way; you simply have to easily enumerate the possibilities for each by treating the roll of each die as independent events.

There are 6 feasible methods to obtain 7 - one for each outcome of the initially die - and two possible ways to obtain 11 - one each in the occasion that the first die is 5 or 6 - interpretation you have actually eight total possibilities . Tbelow are \$6^2=36\$ possibilities for just how the two dice can roll, so you have actually a \$frac836=frac29\$ opportunity of rolling either one.

See more: What Is A Levin Tube Used For Gastric Decompression, What Is A Levin Tube Used For

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answered Aug 14 "20 at 2:26
Stephen GoreeStephen Goree