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Probability: Another Dice Question: Solving Question

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by KevinLuk » Mon Aug 24, 2015 6:40 pm
Assume that each die has 6 sides with faces numbered 1 to 6.

What is the probability that the sum of two dice will yield a 10 or lower?


I'm having a problem with the way it is explained. Thanks.
[spoiler]

The way that I am approaching these questions is that the question is asking for probabilities. The possible outcomes (to me) in which the sums of the dice are over 10 are "5 - 6", "6 - 6", "6 - 6", and "6 - 5". Can somebody please explain to me why the second pair of 6 - 6 is not included in the OA Explanation while "5 - 6" and "6 - 5" are both considered different acceptable probabilities?

Here is the book's explanation:

Solve this problem by calculating the probability that the sum will be higher than 10, and
subtract the result from 1. There are 3 combinations of 2 dice that yield a sum higher than 10: 5 + 6,
6 + 5, and 6 + 6. Therefore, the probability that the sum will be higher than 10 is 3/36, or 1/12. The
probability that the sum will be 10 or lower is 1 - 1/12 = 11/12. [/spoiler]
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by DavidG@VeritasPrep » Mon Aug 24, 2015 7:06 pm
Can somebody please explain to me why the second pair of 6 - 6 is not included in the OA Explanation while "5 - 6" and "6 - 5" are both considered different acceptable probabilities?
The key is that we're talking about different dice. To conceptualize this, Imagine that the two dice are different colors, one blue and one red. The scenarios that will give us a value over 10 are:

A) Red-5/Blue-6
B) Blue-5/Red-6
C) Red-6/Blue-6

A and B are unique scenarios because we're altering which die is turning up '5' and which is turning up '6.' But clearly, Red-6/Blue-6 would be no different from Blue-6/Red-6.
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by Max@Math Revolution » Wed Aug 26, 2015 6:47 am
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


Assume that each die has 6 sides with faces numbered 1 to 6.

What is the probability that the sum of two dice will yield a 10 or lower?


I'm having a problem with the way it is explained. Thanks.

==> the probability that the sum will be smaller than, or will equal 10 = 1 - probability that the sum will be greater than 10

probability = number of cases of a specific event / total number of cases

since two dices have six numbers each, the total number of cases is 6*6=36
the case where the sum is greater than 10 is (5,6),(6,5),(6,6), 3 cases.
therefore the probability that the sum will be smaller than, or will equal 10 = 1-(3C1/36C1)=1-(3/36)=1-(1/12)=11/12



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by [email protected] » Wed Aug 26, 2015 8:57 am
Hi KevinLuk,

You might find it beneficial to 'write everything out' - while it would take a little bit of work, you would be able to physically see everything, which would make dealing with the concept easier.

Since we are rolling two 6-sided dice, there are (6)(6) = 36 possible outcomes. Those outcomes are....

11,12,13,14,15,16
21,22,23,24,25,26
31,32,33,34,35,36
41,42,43,44,45,46
51,52,53,54,55,56
61,62,63,65,65,66

Of these 36 outcomes you can either calculate what you WANT or what you DON'T WANT (and subtract that from the number 1). The second method is easier here...

There are 3 options that DO NOT fit what you're looking for.... 3/36 = 1/12

[spoiler]1 - 1/12 = = 11/12 [/spoiler]= the options that you ARE looking for.

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