What"s the probability of getting a full of \$7\$ or \$11\$ once a pair of fair dice is tossed?

I already looked it up on the internet and also my answer suitable the very same answer ~ above a site. However, though ns am confident the my systems is right, i am curious if there"s a an approach in i beg your pardon I could compute this faster since the photo below shows just how time spend that type of method would be. Thanks in advance.

You are watching: What is the probability of getting a total of seven when you roll two dice   For \$7\$, view that the first roll doesn"t matter. Why? If we roll anything from \$1\$ come \$6\$, then the 2nd roll can always get a sum of \$7\$. The second dice has probability \$frac16\$ the it matches with the an initial roll.

Then, for \$11\$, I choose to think of it together the probability of roll a \$3\$. It"s much easier. Why? shot inverting all the number in your dice table you had actually in the image. Instead of \$1, 2, 3, 4, 5, 6\$, walk \$6, 5, 4, 3, 2, 1\$. You need to see that \$11\$ and also \$3\$ overlap. Indigenous here, just calculate that there space \$2\$ means to role a \$3\$: one of two people \$1, 2\$ or \$2, 1\$. Therefore it"s \$frac236 = frac118\$.

Key takeaways:

\$7\$ is constantly \$frac16\$ probabilityWhen asked to find probability that a larger number (like \$11\$), uncover the smaller counterpart (in this case, \$3\$).
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answered Aug 14 "20 in ~ 2:33 FruDeFruDe
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To calculation the possibility of rolling a \$7\$, role the dice one at a time. Notification that it doesn"t matter what the an initial roll is. Everything it is, there"s one possible roll the the 2nd die that gives you a \$7\$. For this reason the possibility of rolling a \$7\$ needs to be \$frac 16\$.

To calculate the chance of rolling an \$11\$, roll the dice one at a time. If the an initial roll is \$4\$ or less, you have no chance. The very first roll will be \$5\$ or more, maintaining you in the ball game, through probability \$frac 13\$. If you"re quiet in the ball game, your opportunity of acquiring the 2nd roll you need for one \$11\$ is again \$frac 16\$, therefore the complete chance the you will roll one \$11\$ is \$frac 13 cdot frac 16 = frac118\$.

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

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

The variety of positive integer services to \$a_1+a_2=7\$ is \$inom7-12-1=6\$. As such the probability of acquiring \$7\$ from 2 dice is \$frac636=frac16\$.

For \$11\$ or any kind of number greater than \$7\$, us cannot proceed exactly like this, since \$1+10=11\$ is likewise a solution for example, and also we understand that each roll cannot produce greater number 보다 \$6\$. So us modify the equation a tiny to be \$7-a_1+7-a_2=11\$ wherein each \$a\$ is much less than 7. This is identical to finding the variety of positive integers equipment to \$a_1+a_2=3\$, which is \$inom3-12-1=2\$. Therefore, the probability of getting \$11\$ from 2 dice is \$frac236=frac118\$

Try come experiment with various numbers, calculation manually and also using various other methods, then compare the result.

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reply Aug 14 "20 in ~ 2:33 \$endgroup\$
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Welcome to the smashville247.netematics Stack Exchange.

There certain is a quicker way; friend just have actually to quickly enumerate the possibilities for each by treating the roll of each die as independent events.

There room six possible ways to get 7 - one because that each outcome of the very first die - and two feasible ways to gain 11 - one every in the event that the first die is 5 or 6 - meaning you have actually eight complete possibilities . There room \$6^2=36\$ possibilities for just how the 2 dice can roll, therefore you have a \$frac836=frac29\$ chance of rolling one of two people one.

See more: What Happens When You Remove An Electron From An Atom ? Can You Remove Protons From An Atom

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