Four persons can hit a target correctly with probabilities $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$ and $$\frac{1}{8}$$ respectively. If all hit at the target independently, then the probability that the target would be hit, is:

Let us denote the probabilities of the four persons hitting the target by $$p_1=\dfrac12,\;p_2=\dfrac13,\;p_3=\dfrac14,\;p_4=\dfrac18.$$

We want the probability that the target is hit by at least one of them. A convenient way to calculate this is to first find the probability that none of them hits the target, and then subtract that value from 1.

The probability that a person misses the target is obtained by subtracting the probability of a hit from 1. Thus we have

$$q_1 = 1 - p_1 = 1 - \dfrac12 = \dfrac12,$$

$$q_2 = 1 - p_2 = 1 - \dfrac13 = \dfrac23,$$

$$q_3 = 1 - p_3 = 1 - \dfrac14 = \dfrac34,$$

$$q_4 = 1 - p_4 = 1 - \dfrac18 = \dfrac78.$$

Because the shots are independent, the probability that all four persons miss the target is the product of their individual miss-probabilities:

$$P(\text{all miss}) \;=\; q_1 \times q_2 \times q_3 \times q_4 \;=\; \dfrac12 \times \dfrac23 \times \dfrac34 \times \dfrac78.$$

Step-by-step multiplication gives

$$\dfrac12 \times \dfrac23 = \dfrac{1}{3},$$

$$\dfrac{1}{3} \times \dfrac34 = \dfrac14,$$

$$\dfrac14 \times \dfrac78 = \dfrac{7}{32}.$$

So we have

$$P(\text{all miss}) = \dfrac{7}{32}.$$

Now we apply the complementary probability formula

$$P(\text{at least one hit}) = 1 - P(\text{all miss}).$$

Substituting the value just obtained,

$$P(\text{at least one hit}) = 1 - \dfrac{7}{32} = \dfrac{32}{32} - \dfrac{7}{32} = \dfrac{25}{32}.$$

Hence, the correct answer is Option D.