# 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:

$c={b}_{1}+{b}_{2}=0.398+0.504=\frac{451}{500}=0.902$

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