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

CentsLord May 6, 2020

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

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

**\(7775 / 7776\)**

LuckyDucky May 6, 2020

#2**+1 **

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

CentsLord May 6, 2020

#3**+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

#7**+4 **

lol great job

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

hugomimihu
May 6, 2020