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: Determine individual probabilities of hitting the target.
The probabilities of hitting the target are: A at \( \frac{3}{5} \), B at \( \frac{2}{5} \), and C at \( \frac{3}{4} \).Step 2: Determine individual probabilities of missing the target.
The probabilities of missing the target are: A at \( 1 - \frac{3}{5} = \frac{2}{5} \), B at \( 1 - \frac{2}{5} = \frac{3}{5} \), and C at \( 1 - \frac{3}{4} = \frac{1}{4} \).Step 3: Calculate the probability of exactly one successful hit.
This is calculated by summing the probabilities of each person hitting while the others miss:\[P(\text{exactly 1 hit}) = P(\text{A hits, B misses, C misses}) + P(\text{A misses, B hits, C misses}) + P(\text{A misses, B misses, C hits})\]After computation, the probability of exactly one hit yields a final probability for at least two hits of \( \frac{63}{100} \). Final Answer: \[ \boxed{\frac{63}{100}} \]