Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are $$\frac{3}{4}$$, $$\frac{1}{2}$$ and $$\frac{5}{8}$$ respectively, then the probability that the target is hit by P or Q but not by R is:

The probability that person P hits the target is given to be $$\dfrac{3}{4}$$.

Similarly, the probability that person Q hits the target is $$\dfrac{1}{2}$$, and the probability that person R hits the target is $$\dfrac{5}{8}$$.

All three persons act independently, so the probabilities of their individual successes or failures can be multiplied when we need a combined (joint) probability.

We are asked to find the probability that the target is hit by P or Q, but not by R.

First, we consider R’s failure, because the phrase “but not by R” makes R’s missing the target a compulsory condition.

The probability that R misses the target is the complement of the probability that R hits it.

Using the complementary rule $$P(\text{miss}) = 1 - P(\text{hit})$$, we have

$$P(\text{R misses}) = 1 - \dfrac{5}{8} = \dfrac{3}{8}.$$

Next, we need the probability that the target is hit by at least one of P or Q.

Again we use the complementary rule. It is easier to find the probability that neither P nor Q hits, and then subtract that from 1.

The probability that P misses is $$1 - \dfrac{3}{4} = \dfrac{1}{4},$$

and the probability that Q misses is $$1 - \dfrac{1}{2} = \dfrac{1}{2}.$$

Because P and Q act independently, the probability that both miss is the product

$$P(\text{P misses and Q misses}) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}.$$

Therefore, the probability that at least one of P or Q hits is

$$P(\text{P or Q (at least one) hits}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}.$$

Now we combine the two independent requirements:

(i) R must miss, which occurs with probability $$\dfrac{3}{8},$$ and

(ii) at least one of P or Q must hit, which occurs with probability $$\dfrac{7}{8}.$$

Multiplying these independent probabilities, we obtain

$$P(\text{desired event}) = \dfrac{3}{8} \times \dfrac{7}{8} = \dfrac{21}{64}.$$

Hence, the correct answer is Option B.