Originally Posted by

**Grandad** Hello fobos3I assume that the question means that they both shoot together, and continue to shoot at the target until at least one of them hits it. At this point we want to know the probabilities that (i) they both hit at the same time; (ii) one hits and the other misses. Is that how you understand the question?

If that is so, then I would argue as follows:

On any given turn, the probability that they both hit is $\displaystyle 0.2 \times 0.2 = 0.04$ and the probability that one hits and one misses is $\displaystyle 2 \times 0.2 \times 0.8 = 0.32$. So, on any given occasion, the probability that one hit and one miss occurs is $\displaystyle \frac{0.32}{0.04} = 8$ times greater than the probability that two hits occur.

Now if my assumption about the rules of the game is correct, we can ignore all probabilities until a hit is achieved. But when a hit occurs, the probability that it has been achieved because both have scored a hit is $\displaystyle \frac{1}{9}$ and the probability that one has hit and the other has missed is $\displaystyle \frac{8}{9}$.

Grandad