Question

A can hit a target 3 times in 5 shots, B 2 times in 5 shots, and C three times in 4 shots. All of them fire one shot each simultaneously at the target. What is the probability that at least two shots hit?

• A. \( \frac{63}{100} \)
• B. \( \frac{9}{20} \)
• C. \( \frac{98}{20825} \)
• D. \( \frac{396}{10025} \)

Hint:

When calculating probabilities for multiple independent events, use the rule for the union of probabilities and account for all possible combinations of hits and misses.

Answer 1

The Correct Option is A

Step 1: Calculate the probability of hitting the target.
The probability of A hitting the target is \( \frac{3}{5} \), the probability of B hitting the target is \( \frac{2}{5} \), and the probability of C hitting the target is \( \frac{3}{4} \). Step 2: Find the probability of missing the target.
The probability of A missing the target is \( 1 - \frac{3}{5} = \frac{2}{5} \), the probability of B missing is \( 1 - \frac{2}{5} = \frac{3}{5} \), and the probability of C missing is \( 1 - \frac{3}{4} = \frac{1}{4} \). Step 3: Calculate the probability of exactly one hit.
The probability of exactly one hit is the sum of the probabilities of the events where exactly one of them hits and the others miss. \[ P(\text{exactly 1 hit}) = P(\text{A hits, B misses, C misses}) + P(\text{A misses, B hits, C misses}) + P(\text{A misses, B misses, C hits}) \] This can be computed, and the final probability for at least two hits is \( \frac{63}{100} \). Final Answer: \[ \boxed{\frac{63}{100}} \]