Hi! I believe the clearest way to show you how to solve this easily is to do a tree diagram.

Since the man can only hit the target once in four shots, we assume that the probability of him hitting the target is 0.25. Conversely, the probability of him not hitting the target is 3 in four shots or 0.75.

You want to find out the probability of him landing a shot. This means that for all four shots in succession that he takes, he could hit once, twice, thrice or all four times. He could miss once then hit or miss twice and then hit. There are many possibilities. Too many to slowly add. Therefore, to simplify this, it would much easier to calculate the reverse which is the probability that he doesn't land a shot at all. Any other probability would include him managing to land a shot.

Using the tree diagram, you can see that the branches split off into hit and don't hit. I opted not to show the entire tree diagram as it would be an unnecessary endeavour. I don't need to find out if he hits the target. I only need to follow the branch where he doesn't hit the target at all. The only branch that follows this though is the last one where he misses all four shots.

Hence, the probability that he does not hit the target at all is 0.75 multiplied by itself four times in a row as he misses all four shots. From this equation, we get 0.31640625 which is the probability of him not landing a shot at all.

Ultimately, to find the probability that he lands a shot, we take the total probability (1) and minus off the probability that he does not land a shot. The last branch of the tree is cut off.

We then get 0.68. This is the probability that he lands at least a shot and there you have your answer.

You can try drawing out the rest of the branches for better clarification.

I hope this helped. All the best!