Hey folks, thanks for your hints and answers!
I want to point out here three things, first the original question, second my approach, third a question related to multinomial usage in my approach.
1.) The original question comes from Schaums, I quote it here: "Find the probability of not getting a 1,2 or 3 in 4 tosses of a fair die". A solution without a way is also given with 1/16. The headline is "multinomial distribution", so it should be solved with this...
2.) My approach is concerning, that "not getting a 1,2 or 3 in 4 tosses" means, I can get 4,5,6. But the question is however which pattern this means e.g. 4XXX, 44XXX and so on. I was assuming that the following patterns are allowed, for which I also calculated the probability:
- Pattern 4444, 5555, 6666 (4 times same number) --> [MATH]3\left(\frac{1}{6}\right)^4[/MATH]- Pattern 4455, 4466 , ... (each to numbers are equal) --> [MATH]3\cdot \frac{4!}{2!2!}\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^2[/MATH]- Pattern 4456, 5546, ... (two numbers equal, and two different) --> [MATH]3\left(\frac{4!}{2!1!1!}\right)\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^1[/MATH]- Pattern 444(5|6), ... (three equal numbers and one of two different) --> [MATH]6\left(\frac{4!}{3!1!}\right)\left(\frac{1}{6}\right)^3\left(\frac{1}{6}\right)^1[/MATH]
Putting it all together:
[MATH]3\left(\frac{1}{6}\right)^4+ 3\cdot \frac{4!}{2!2!}\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^2 + 3\left(\frac{4!}{2!1!1!}\right)\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^1 + 6\left(\frac{4!}{3!1!}\right)\left(\frac{1}{6}\right)^3\left(\frac{1}{6}\right)^1 = \frac{1}{16}[/MATH]
3.) Now as I see my calculation it "looks" like multinomial distribution, but I'am not sure if so, therefore I would be very thankful, if someone can confirm if this is the right approach