Three shooters fire one shot, hitting probability 0.5; 0.6; 0.8. Find the probability that at least one of them will miss.

1. Let A1, A2 and A3 be the events when each of the three shooters, respectively, hits the target:

P (A1) = 0.5;
P (A2) = 0.6;
P (A3) = 0.8.
2. Since these events are independent, the probability of event X that all three arrows hit the target is equal to the product of the probabilities of each of the three events:

P (X) = P (A1) * P (A2) * P (A3);
P (X) = 0.5 * 0.6 * 0.8 = 0.24.
3. The opposite event Y is that not all three arrows hit the target, that is, at least one of them will miss. Therefore, the probability of this event is:

P (Y) = 1 – P (X) = 1 – 0.24 = 0.76.
Answer: 0.76.