A can hit a target 4 times in 5 shots, B 3 times in 4 shots; C twice in 3 shots. They fire atarget. What is the probability that at least two shots hit?

To solve this problem, we need to consider all the possible ways at least two shots can hit the target. This includes the scenarios where exactly two shots hit, exactly three shots hit, and all three shots hit.

1. Exactly two shots hit:

   • A and B hit, C misses. The probability is (4/5) * (3/4) * (1/3) = 0.2
   • A and C hit, B misses. The probability is (4/5) * (1/4) * (2/3) = 0.1333
   • B and C hit, A misses. The probability is (1/5) * (3/4) * (2/3) = 0.1 The total probability for exactly two shots hit is 0.2 + 0.1333 + 0.1 = 0.4333

2. Exactly three shots hit:

   • All three hit. The probability is (4/5) * (3/4) * (2/3) = 0.4

3. All three shots hit:

   • The probability is the same as the scenario where exactly three shots hit, which is 0.4

So, the total probability that at least two shots hit is the sum of the probabilities for the three scenarios, which is 0.4333 + 0.4 = 0.8333 or approximately 83.33%.