The probability of hitting the target by the first shooter = 0.8 by the second shooter = 0.6, and by the third 0.5.

The probability of hitting the target by the first shooter = 0.8 by the second shooter = 0.6, and by the third 0.5. Find the probability that at least one shooter will hit the target.

Hitting the target with different arrows are independent events.
Missing probabilities for each of the shooters:
q1 = 1 – 0.8 = 0.2; q2 = 1 – 0.6 = 0.4; q3 = 1 – 0.5 = 0.5;
The probability that there will be three misses:
P = q1 * q2 * q3 = 0.2 * 0.4 * 0.5 = 0.04;
The probability that the opposite event will occur, in which at least one shooter hits the target:
P` = 1 – P = 1 – 0.04 = 0.96.
Answer: 0.96.