So, I'm playing a board game in which there are 5 dice (which you roll at once), 3 ships and a probability of hitting one ship with one die of 2/6.
My question: What is the probability of hitting all 3 ships using the 5 dice?
So far I've got this: $$P=\biggl(\frac 2 6\biggr)^3 + \biggl(1-\frac 35\biggr)\biggl(\frac 2 6\biggr)^3 = 0.051852 $$
Is it good?
Presumably your $\left( \frac 26\right)^3$ is the chance the first three dice come up hits. I don't understand the logic of the $\left(1-\frac 35\right)$ You can either count the chance you get exactly $3,4,5$ and add or count the chance you get $0,1,2$ and subtract from $1$. To get $5$ hits is $\left( \frac 26\right)^5$. To get three is ${5 \choose 3}\left( \frac 26\right)^3\left( \frac 46\right)^2$ because you choose which three dice will hit, those come up right $\frac 26$ of the time and the others have to miss.