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

Step-by-step explanation:

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

answer: f = 32.2m

step-by-step explanation: that is the answer