A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are $$\frac{3}{4}$$, $$\frac{1}{2}$$, $$\frac{5}{8}$$. The probability that the target is hit by A or B but not by C is :

We are given the probabilities of A, B, and C hitting the target independently:

• Probability that A hits, P(A) = $$\frac{3}{4}$$
• Probability that B hits, P(B) = $$\frac{1}{2}$$
• Probability that C hits, P(C) = $$\frac{5}{8}$$

We need the probability that the target is hit by A or B but not by C. This means at least one of A or B hits the target, and C does not hit. We can write this event as (A ∪ B) ∩ C'. Since A, B, and C are independent, the events A ∪ B and C' are also independent. Therefore, we can compute:

P((A ∪ B) ∩ C') = P(A ∪ B) × P(C')

First, find P(C'), the probability that C does not hit:

P(C') = 1 - P(C) = 1 - $$\frac{5}{8}$$ = $$\frac{8}{8}$$ - $$\frac{5}{8}$$ = $$\frac{3}{8}$$

Next, find P(A ∪ B), the probability that at least one of A or B hits. Using the inclusion-exclusion principle:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Since A and B are independent, P(A ∩ B) = P(A) × P(B):

P(A ∩ B) = $$\frac{3}{4}$$ × $$\frac{1}{2}$$ = $$\frac{3}{8}$$

Now substitute:

P(A ∪ B) = $$\frac{3}{4}$$ + $$\frac{1}{2}$$ - $$\frac{3}{8}$$

Convert to a common denominator of 8:

$$\frac{3}{4}$$ = $$\frac{3 \times 2}{4 \times 2}$$ = $$\frac{6}{8}$$

$$\frac{1}{2}$$ = $$\frac{1 \times 4}{2 \times 4}$$ = $$\frac{4}{8}$$

So, P(A ∪ B) = $$\frac{6}{8}$$ + $$\frac{4}{8}$$ - $$\frac{3}{8}$$ = $$\frac{6 + 4 - 3}{8}$$ = $$\frac{7}{8}$$

Now, compute P((A ∪ B) ∩ C') = P(A ∪ B) × P(C') = $$\frac{7}{8}$$ × $$\frac{3}{8}$$ = $$\frac{7 \times 3}{8 \times 8}$$ = $$\frac{21}{64}$$

To verify, we can break the event into mutually exclusive cases:

1. A hits, B misses, C misses: P(A) × P(B') × P(C')
2. A misses, B hits, C misses: P(A') × P(B) × P(C')
3. A hits, B hits, C misses: P(A) × P(B) × P(C')

Find the probabilities:

P(B') = 1 - P(B) = 1 - $$\frac{1}{2}$$ = $$\frac{1}{2}$$

P(A') = 1 - P(A) = 1 - $$\frac{3}{4}$$ = $$\frac{1}{4}$$

P(C') = $$\frac{3}{8}$$ (as before)

Case 1: $$\frac{3}{4}$$ × $$\frac{1}{2}$$ × $$\frac{3}{8}$$ = $$\frac{3 \times 1 \times 3}{4 \times 2 \times 8}$$ = $$\frac{9}{64}$$

Case 2: $$\frac{1}{4}$$ × $$\frac{1}{2}$$ × $$\frac{3}{8}$$ = $$\frac{1 \times 1 \times 3}{4 \times 2 \times 8}$$ = $$\frac{3}{64}$$

Case 3: $$\frac{3}{4}$$ × $$\frac{1}{2}$$ × $$\frac{3}{8}$$ = $$\frac{3 \times 1 \times 3}{4 \times 2 \times 8}$$ = $$\frac{9}{64}$$

Sum the probabilities: $$\frac{9}{64}$$ + $$\frac{3}{64}$$ + $$\frac{9}{64}$$ = $$\frac{9 + 3 + 9}{64}$$ = $$\frac{21}{64}$$

Both methods give the same result. Therefore, the probability is $$\frac{21}{64}$$.

Hence, the correct answer is Option A.