+770 What is the probability that when we roll 5 fair 6-sided dice, at most 4 of them will show a 1?
+639 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 = \(1 - (1/6)^5 \)
\((6^5 - 1) / 6^5 =\)
\(7775 / 7776\)
+770 Why is the probability that at most 4 will show one =\(1 - (\frac{1}{6})^2\)?
+1008 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
+1008 lol great job
I mastered all in intro counting probability and weeks 1-5 geometry lol
