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Basic Dice Probability Question

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Basic Dice Probability Question

by hotwann » Mon Aug 27, 2012 4:06 pm
Hi - I had question regarding probabilities with a six-sided die.

The MGMAT Number Properties study guide has the following example question:
What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?
So, after reading, I understand that you could calculate this by finding that for each of the three rolls, there is a 5/6 probability that the die will not yield a 6. Or, you can do 1/6+5/6*1/6+5/6*5/6*1/6 to add the probabilities of each of the three scenarios.

So, for some reason, my dumb brain naturally wants to solve this problem by just saying "1/6 + 1/6 + 1/6."

I understand it is not the answer to the question, given that it doesn't not account for the three scenarios correctly. But conceptually, what would 1/6 + 1/6 + 1/6 represent? Is there a way to put that into words? That might help me understand better why it's wrong.

Thanks
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by SmartAssJun » Mon Aug 27, 2012 4:59 pm
hotwann wrote:Hi - I had question regarding probabilities with a six-sided die.

The MGMAT Number Properties study guide has the following example question:
What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?
So, after reading, I understand that you could calculate this by finding that for each of the three rolls, there is a 5/6 probability that the die will not yield a 6. Or, you can do 1/6+5/6*1/6+5/6*5/6*1/6 to add the probabilities of each of the three scenarios.

So, for some reason, my dumb brain naturally wants to solve this problem by just saying "1/6 + 1/6 + 1/6."

I understand it is not the answer to the question, given that it doesn't not account for the three scenarios correctly. But conceptually, what would 1/6 + 1/6 + 1/6 represent? Is there a way to put that into words? That might help me understand better why it's wrong.

Thanks
First let's get the probability that none of the rolls are 6.
The probablility of each roll of the dice not being 6 is 5/6
So the probabilty of none of the rolls being 6 should be (5/6)^3
which is 125/216. That's what the probability of that is.
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by SmartAssJun » Mon Aug 27, 2012 5:03 pm
SmartAssJun wrote:
hotwann wrote:Hi - I had question regarding probabilities with a six-sided die.

The MGMAT Number Properties study guide has the following example question:
What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?
So, after reading, I understand that you could calculate this by finding that for each of the three rolls, there is a 5/6 probability that the die will not yield a 6. Or, you can do 1/6+5/6*1/6+5/6*5/6*1/6 to add the probabilities of each of the three scenarios.

So, for some reason, my dumb brain naturally wants to solve this problem by just saying "1/6 + 1/6 + 1/6."

I understand it is not the answer to the question, given that it doesn't not account for the three scenarios correctly. But conceptually, what would 1/6 + 1/6 + 1/6 represent? Is there a way to put that into words? That might help me understand better why it's wrong.

Thanks
First let's get the probability that none of the rolls are 6.
The probablility of each roll of the dice not being 6 is 5/6
So the probabilty of none of the rolls being 6 should be (5/6)^3
which is 125/216. So the probability that at least one of the rolls is 6
is 1-(125/216) = 91/216, that's how possibily you'd get 6 at least once out of
three rolls.
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by mehaksal » Tue Aug 28, 2012 3:05 am
shouldn't the answer to this question be :-
1/6*5/6*5/6 + 1/6*/1/6*5/6 + 1/6*1/6*1/6
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by Ian Stewart » Tue Aug 28, 2012 5:49 pm
hotwann wrote: So, for some reason, my dumb brain naturally wants to solve this problem by just saying "1/6 + 1/6 + 1/6."

I understand it is not the answer to the question, given that it doesn't not account for the three scenarios correctly. But conceptually, what would 1/6 + 1/6 + 1/6 represent? Is there a way to put that into words? That might help me understand better why it's wrong.
In probability (and more generally in counting problems, for the same reasons), we add when we have different cases. So if you were asked "what is the probability when you roll a die once that you get a 1, 2 or 5?" then one (long) way to answer the problem is to break it into cases. We can get the result we want in one of three distinct ways: we roll a 1 *or* we roll a 2 *or* we roll a 5. We can add the probability of each case to get the answer: 1/6 + 1/6 + 1/6 = 3/6. So that's the type of situation where you might arrive at the type of answer you wanted to write down here - where there are three distinct ways to get the result the question asks for, and each has a probability of 1/6.

Notice that we do this kind of addition in daily life all the time too - if you know the probability of rain is 40% and the probability of snow is 15%, then (assuming it can't both rain and snow) you would probably quite naturally say that the probability of rain or snow would be 55%, and that would be perfectly correct.

Now, you multiply in probability (and, more generally, in counting) in a completely different situation: when you need a sequence of results to all happen. So if you roll a die three times, and want to know the probability you get a '6' on each roll, you would multiply the probability that happens each time: 1/6 * 1/6 * 1/6 = 1/6^3. So if you ever see a probability problem where you're rolling a die three times, or flipping a coin three times, or whatever, you're usually going to end up multiplying at some point because you're going to have to consider a sequence of three outcomes.

And, of course in more complicated questions, the two situations described above can be combined, so you may need to both add and multiply in the same question - e.g. 'if you flip a coin 3 times, what's the probability you get either 3 heads or 3 tails?'. Then you have two cases (all heads, or all tails), but to find the probability of each case, we'll need to multiply.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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