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by Jay@ManhattanReview » Wed Dec 14, 2016 4:18 am
matt1234 wrote:you roll a die 5 times, what is the probability that you get a 6 exactly once


Please help thank you!!!!
The probability of getting a '6' = 1/6;

=> The probability of NOT getting a '6' = 5/6; (we can get any number other than the '6': 1, 2, 3, 4, or 5

=> (Probability of getting a '6' on the first throw) * (Probability of NOT getting a '6' on the subsequent four rolls) = (1/6)*(5/6)*(5/6)*(5/6)*(5/6)

= (1/6)*(5/6)^4

= 5^4/6^5

The '6' not necessarily should show up on the first roll, it can show up on 2nd, 3rd, 4th, or 5th roll too.

=> Probability that you get a 6 exactly once = 5C1 * (5^4/6^5) = 5 * 5^4/6^5 = (5/6)^5

Hope this helps!

-Jay

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by [email protected] » Wed Dec 14, 2016 10:11 am
Hi matt1234,

In these types of prompts, if you don't immediately 'see' the elegant approach to getting to the correct answer, it often helps to 'map' out what you're looking to calculate. In that way, you might be able to discover a short-cut that can save you some time as you're working through the math.

We're asked to roll a die 5 times and we're asked for the probability of getting EXACTLY ONE 6. Let's start by mapping out one possible option:

First roll is 6, the last four rolls are 'not 6': (1/6)(5/6)(5/6)(5/6)(5/6) = (5^4)/(6^5)

Of course, we could end up getting the lone 6 on the SECOND roll, with the other 4 rolls being 'not 6'; that would look like this....

(5/6)(1/6)(5/6)(5/6)(5/6) = (5^4)/(6^5)

Notice how the outcome is exactly the SAME. This proves that we don't have to redo any additional calculation. There are 5 'spots', so there are 5 options to end up with just one 6. Thus, we just have to multiply the above calculation by 5....

(5)(5^4)/(6^5)

The numerator can then be 'combined'....

(5^5)/(6^5)

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by Scott@TargetTestPrep » Thu Dec 15, 2016 4:50 pm
matt1234 wrote:you roll a die 5 times, what is the probability that you get a 6 exactly once
The probability of rolling a 6 is 1/6, and the probability of NOT rolling a 6 is 1 - 1/6 = 5/6.

So, if we are rolling a 6 exactly once in 5 tries, we are rolling a 6 in one of the 5 tries and a number other than 6 in the other 4 tries. We can calculate this as follows (note: S = rolling a 6, NS = not rolling a 6):

P(S x NS x NS x NS x NS) = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 = 625/7776

However, there are 5 ways in which we could roll exactly one 6:

S-NS-NS-NS-NS

NS-S-NS-NS-NS

NS-NS-S-NS-NS

NS-NS-NS-S-NS

NS-NS-NS-NS-S

Thus, the final probability of rolling just one 6 is 625/7776 x 5 = 3,125/7,776

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by crackverbal » Thu Dec 15, 2016 8:56 pm
Hi Matt,

Pretty much everyone has posted what could be different ways of approaching this question.

I will give my 2 cents to this question.

These types of question can be solved using combinatorial approach, but the below rule would help you to approach it better.

Probability of Complex event = Product of individual events * arrangements

Here its rolling the die five times and getting six exactly once.

So according to the above rule it is, (six)(Not six) (Not six) (Not six) (Not six) * arrangement

P(getting six) = 1/6

P(not getting six) = 5/6

So, 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * arrangement.

Arrangement can be done many ways, already few have posted different ways (like selections, listing..) of arranging.

Another alternative way is,

Consider this as a word,

SNNNN,

S - Six , N - Not Six

So SNNNN can be arranged in 5!/4! Ways. That is 5 different ways.

If you are not aware of the above arrangement,

Let's say for an example, Given a word AISHA - different ways of arranging it is 5!/ 2!. Here the numerator is
arranging of 5 letters and denominator is dividing the repetitions.

So the final answer would be 5^ 5/ 6^5.

So if you look at it finding the probability is similar but arrangements can be done in multiple ways.

So you can chose best way you are comfortable with.

Hope this is clear.
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