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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
avatar+657 
0

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 = \(1 - (1/6)^5 \)

 

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

 

\(7775 / 7776\)

.
 May 6, 2020
 #2
avatar+300 
+1

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

 May 6, 2020
 #3
avatar+1498 
+4

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
avatar+300 
+1

Thank you! :D Got it right

CentsLord  May 6, 2020
 #5
avatar+1498 
+2

np glad to help :))))

 

laugh

hugomimihu  May 6, 2020
 #6
avatar+657 
+2

Yo CentLord I just got mastered in every alcumus topic week 1-7 YAY!

LuckyDucky  May 6, 2020
 #7
avatar+1498 
+4

lol great job

 

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

hugomimihu  May 6, 2020
 #8
avatar+657 
+1

LOL nice!

LuckyDucky  May 6, 2020

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