5 regular 6 sided dice are thrown. What is the probability to get at least 4 dice with 6:es?
The "at least" part is a bit tricky.
Solved it.
Found this video that helped me solve it.
link
"At least four" means "four or five". I don't see why that would be "tricky".
Write "S" to mean a six, "N" to mean "not a six". Then one way of getting 4 sixes in five rolls would be "SSSSN", another would be "SSSNS", etc. It should be easy to see that there are five ways (NOT 4) to write those: the single "N" in each of the 5 places. And the probability of each of those is $\displaystyle \left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^4$ so the probability of "four sixes in five rolls" is $\displaystyle 5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^ 4$.
The probability of 5 sixes in 5 rolls is, of course, $\displaystyle \left(\frac{1}{6}\right)^5$. The probability of "at least four sixes= either 4 sixes or 5 sixes" is the sum of those:
$\displaystyle 5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^ 4+ \left(\frac{1}{6}\right)^5$.
(Again, "5", not "4". That could also be calculated as $\displaystyle \frac{5!}{4! 1!}$.)
Yes, thanks for the clarifications, it's a lot easier to recognize the path to solution when the technique is clarified.
And the technique is basically:
First step is identifying where the at least bit is, in my case it was at 4 out of 5.
Next step is to sum all remaining bits, including the staring one, with the following formula:
P = n C r * p^r * q^(n-r) ~ as explained in the video.