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Re: Probability of at least one outcome [#permalink]
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rooker wrote:
Thanks Zarrolou.

I'm having trouble understanding the concept of probability.
Im trying to frame this question as a Combinatorics question. Please help me out.


Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be : What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1 AND 5 AND 5 OR 5 AND 1 AND 5 OR 5 AND 5 AND 1.

That would be 75.

So, probability of getting exactly one 6 would now be, \(\frac{Ways of getting desired outcome}{All possible outcomes}\) = \(\frac{75}{216}\)

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.


Absolutely correct! The probability of getting one 6 and three N6s is: \(\frac{1}{6}*\frac{5}{6}*\frac{5}{6}\)
The number of possible outcome is which this can happen is 3. So \(3*\frac{25}{216}\) is the magic number
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Re: Probability of at least one outcome [#permalink]
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Hi there,

This is my way of approach:
The probability of not getting a 6 in 3 throws is (5/6)X(5/6)x(5/6).
Therefore the probability of getting at least one 6 is 1- [P(no 6 in 3 throws) = (5/6)X(5/6)x(5/6)] = 91/216
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Re: What is the probability that, on three rolls of a number cub [#permalink]
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Hi All,

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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Re: What is the probability that, on three rolls of a number cub [#permalink]
Probability of getting at least 1 six in three rolls = 1- (no six in three rolls)

Probability of no six in 3 rolls = (5/6) * (5/6) * (5/6) = (125/216)

Thus, answer = 1- (125/216) = (91/216)
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Re: What is the probability that, on three rolls of a number cub [#permalink]
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Re: What is the probability that, on three rolls of a number cub [#permalink]
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