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Manhattan Strategy Guide - Number Properties

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Manhattan Strategy Guide - Number Properties

by [email protected] » Mon Feb 02, 2015 5:58 pm
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?

The explanation given in the guide is that the probability of not getting a 6 in all 3 rounds is 5/6*5/6*5/6 = 125/216
Therefore, the probability of getting at least one 6 is 1-125/216.

My query is how is the case set up? Meaning, Shouldn't the probability of not getting a 6 on all 3 rounds + getting a 6 on all three rounds = 1. If yes, then how can we subtract 1-125/216.

Please explain.
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by DavidG@VeritasPrep » Mon Feb 02, 2015 6:32 pm
This is just an application of the axiom P(x) = 1 - P(Not x)

If there were a 20% chance of rain, for example, there'd be an 80% chance of no rain. And clearly .2 = (1-.8.)

Now let's transition to dice rolls. Start with a simple case:

Say we just want the probability of rolling a 6 on one roll. Clearly, it's 1/6. So then the probability of not rolling a 6 would be 1 - 1/6 = 5/6.

Now, let's say we roll twice, and we want the probability of rolling at least one six. Well P(at least one 6) = 1 - P(Zero 6's.) So let's calculate the probability of getting zero 6's on two rolls. It would be 5/6* 5/6 = 25/36. Which means that P(at least one 6) = 1 - 25/36 = 11/36.

Same logic if we roll three times and want at least one 6. Again, P(at least one 6) = 1 - P(Zero 6's.) Now we'll calculate the probability of getting zero 6's on three rolls. That would be 5/6 * 5/6 * 5/6 = 125/216. Which means that P(at least one 6) = 1 - 125/216 = 91/216. (And not notice that if you were to add the probability of getting no sixes (125/216) to the probability of getting at least one six (91/216) you would, in fact, get 1.
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by Brent@GMATPrepNow » Mon Feb 02, 2015 6:48 pm
[email protected] wrote: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?

The explanation given in the guide is that the probability of not getting a 6 in all 3 rounds is 5/6*5/6*5/6 = 125/216
Therefore, the probability of getting at least one 6 is 1-125/216.

My query is how is the case set up? Meaning, Shouldn't the probability of not getting a 6 on all 3 rounds + getting a 6 on all three rounds = 1. If yes, then how can we subtract 1-125/216.

Please explain.
Here's another way to look at it:
I offer you a $100 prize if you can roll three dice and NOT get zero 6's.
Does this mean, that you must roll three 6's in order to collect $100?
No.

"NOT getting zero 6's" = "getting one 6" OR "getting two 6's" OR "getting three 6's"
In other words, "NOT getting zero 6's" is the OPPOSITE (aka COMPLEMENT) of "getting at least one 6"

Cheers,
Brent
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by [email protected] » Mon Feb 02, 2015 9:57 pm
Hi nishatfarhat87,

When it comes to probability questions, there are only 2 things that you can figure out:

1) What you WANT to have happen.
2) What you DON'T WANT to have happen.

If you add those two things together, then you get the number 1.

For some probability questions, it's pretty easy to figure out the exact thing that you want to have happen. In other questions though, it's easier to figure out what you DON'T WANT, then subtract that probability from 1 (and you'll have the probability of what you DO want).

In the example that you posted, we're asked for the probability of rolling 3 dice and getting AT LEAST one 6. That means getting one 6, two 6s or three 6s would fit the definition of what we WANT. That seems like a LOT of different possibilities to keep track of. It's easier to figure out what we DON'T WANT (which would be zero 6s).

With 3 dice rolls, the probability of getting zero 6s is....

(not a 6)(not a 6)(not a 6) = (5/6)(5/6)(5/6) = 125/216

1 - 125/216 = 91/216 = the probability of getting AT LEAST ONE 6.

As you deal with additional probability questions during your studies, you have to be clear on what the question asks you to figure out, then think about which of the two approaches would be easier.

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by misterholmes » Thu Feb 05, 2015 1:06 pm
[email protected] wrote: My query is how is the case set up? Meaning, Shouldn't the probability of not getting a 6 on all 3 rounds + getting a 6 on all three rounds = 1.
The setup you propose doesn't account for the cases when you get a 6 on just some of the rounds. It's not a case of all or none.

Allow me to plug my GMAT Quant Diagnostic https://gmatdojo.files.wordpress.com/20 ... nostic.pdf. If you send me your work I can give you some feedback on the next steps.

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by Matt@VeritasPrep » Sun Feb 08, 2015 11:49 pm
There are a few cases here:

No sixes = (5/6) * (5/6) * (5/6)

Exactly one six = (5/6) * (5/6) * (1/6) * 3

Exactly two sixes = (5/6) * (1/6) * (1/6) * 3

Exactly three sixes = (1/6) * (1/6) * (1/6)

The middle two cases are tricky; we multiply by three because there are three arrangements in which the dice could land. (For instance, the first die could be the 6, so the probability of 6-Not 6-Not 6 = (1/6)(5/6)(5/6), but the second die could be the 6, so the probability of Not 6-6-Not 6 = (5/6)(1/6)(5/6), etc.)

As you can see, the easiest way to get the cases we want -- at least one six -- is to lump the last three together, and find them by doing 1 - (the case we don't want), or 1 - (No Sixes).
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