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# How should I approach this?

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What is the probability that when we roll 5 fair 6-sided dice, at most 4 of them will show a 1?

May 6, 2020

#1
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Ok Heres what I would do:

So the probability that all will show a 1 is $$(1/6)^5.$$

The probability that at most 4 will show one = $$1 - (1/6)^5$$

$$(6^5 - 1) / 6^5 =$$

$$7775 / 7776$$

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May 6, 2020
#2
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Why is the probability that at most 4 will show one =$$1 - (\frac{1}{6})^2$$?

May 6, 2020
#3
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it's basically "what's the probability that at least 2 don't roll to be 1?"

So the probability of getting 2 1s is (1/6)^2  and then 1 is total probability so we subtract.

Thus, 1 - 1/36

BUT i think we need to do casework

You see, we have the ones where there's 0 ones, 1 one, ...

then you have to add those too?

idk

hugomimihu  May 6, 2020
#4
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Thank you! :D Got it right

CentsLord  May 6, 2020
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np glad to help :))))

hugomimihu  May 6, 2020
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Yo CentLord I just got mastered in every alcumus topic week 1-7 YAY!

LuckyDucky  May 6, 2020
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lol great job

I mastered all in intro counting probability and weeks 1-5 geometry lol

hugomimihu  May 6, 2020
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LOL nice!

LuckyDucky  May 6, 2020