My problem:
We repeat event A five times independently.
Event A:We throw four dices.
What is the probability that we get exactly two dices showing a "one" none of the five times?
The answer:
P(exactly ⚀⚀ none of the five times) = 0.54
My attempt of solving the problem:
P(exactly ⚀⚀ none of the five times) = 1 - P(exactly ⚀⚀ all of the times)
P(exactly ⚀⚀ one time) = (1/6)(1/6)(5/6)(5/6) = (25/1296) = 0.0193
P(exactly ⚀⚀ all of the times) = (25/1296)5 = 2.6710070610-9
P(exactly ⚀⚀ none of the five times) = 1-2.6710070610-9 = 0.99999999732
My Question:
I don't get the correct answer, what am I doing wrong? I am very thankful for any help and/or guidance.
Your error is in the counter-probability: The counter event is "At least one times we get 1-1", which is more likely than getting "1-1 all five times".
Anyway, I wouldn't use the counter-probability, it's easier to calculate the probability directly:
Of the $6^4$ possibilities we have $\binom{4}{2} \cdot 5^2$ possibilities of having exactly two dice with a $4$: $\binom 4 2$ is needed to choose the two dices that gave a $1$ and the $5^2$ is necessary to get the other possible numbers (besides $1$). Therefore we have:
$$ P(\text{exactly 1-1 none of the five times}) = \left(1-\frac{\binom 4 2 \cdot 5^2}{6^4}\right)^5 \approx 0.884^5 \approx 0.54 $$