Three shooters shoot at the same target, each of them shoots just once. The first one hits the target with a probability of 50%, the second one with a probability of 60% and the third one with a probability of 70%. What is the probability that the shooters will hit the target

1. at least once?2. at least twice?

Let X be the number of times the target is hit. The probability P (X≥1) then equals 1 minus the probability of missing the target three times:

P (X≥1) = 1 - (1-P (A)) (1-P (B)) (1-P (C))

= 1-0.4*0.3*0.2

= 0.976

To find the probability P (X≥2) of hitting the target at least twice, you can consider two cases: either two people hit the target and one does not, or all people hit the target. We find:

P (X≥2) = (0.4*0.7*0.8) + (0.6*0.3*0.8) + (0.6*0.7*0.2) + (0.6*0.7*0.8) = 0.788