Three shooters

Three shooters shoot, each one time, on the same target. The first hit the target with a probability of 0.7; second with a probability of 0.8 and a third with a probability of 0.9. What is the probability to hit the target:

a) just once
b) at least once
c) at least twice

Correct result:

a =  0.092
b =  0.994
c =  0.902

Solution:

p1=0.7 p2=0.8 p3=0.9  n1=1p1=10.7=310=0.3 n2=1p2=10.8=15=0.2 n3=1p3=10.9=110=0.1  a=p1 n2 n3+n1 p2 n3+n1 n2 p3=0.7 0.2 0.1+0.3 0.8 0.1+0.3 0.2 0.9=23250=0.092
b1=p1 p2 n3+n1 p2 p3+p1 n2 p3=0.7 0.8 0.1+0.3 0.8 0.9+0.7 0.2 0.9=199500=0.398 b2=p1 p2 p3=0.7 0.8 0.9=63125=0.504  b=a+b1+b2=0.092+0.398+0.504=497500=0.994
c=b1+b2=0.398+0.504=451500=0.902



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