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

Result

a =  0.092
b =  0.994
c =  0.902

Solution:

$p_{ 1 } = 0.7 \ \\ p_{ 2 } = 0.8 \ \\ p_{ 3 } = 0.9 \ \\ \ \\ n_{ 1 } = 1-p_{ 1 } = 1-0.7 = \dfrac{ 3 }{ 10 } = 0.3 \ \\ n_{ 2 } = 1-p_{ 2 } = 1-0.8 = \dfrac{ 1 }{ 5 } = 0.2 \ \\ n_{ 3 } = 1-p_{ 3 } = 1-0.9 = \dfrac{ 1 }{ 10 } = 0.1 \ \\ \ \\ a = p_{ 1 } \cdot \ n_{ 2 } \cdot \ n_{ 3 }+n_{ 1 } \cdot \ p_{ 2 } \cdot \ n_{ 3 }+n_{ 1 } \cdot \ n_{ 2 } \cdot \ p_{ 3 } = 0.7 \cdot \ 0.2 \cdot \ 0.1+0.3 \cdot \ 0.8 \cdot \ 0.1+0.3 \cdot \ 0.2 \cdot \ 0.9 = \dfrac{ 23 }{ 250 } = 0.092$

$b_{ 1 } = p_{ 1 } \cdot \ p_{ 2 } \cdot \ n_{ 3 }+n_{ 1 } \cdot \ p_{ 2 } \cdot \ p_{ 3 }+p_{ 1 } \cdot \ n_{ 2 } \cdot \ p_{ 3 } = 0.7 \cdot \ 0.8 \cdot \ 0.1+0.3 \cdot \ 0.8 \cdot \ 0.9+0.7 \cdot \ 0.2 \cdot \ 0.9 = \dfrac{ 199 }{ 500 } = 0.398 \ \\ b_{ 2 } = p_{ 1 } \cdot \ p_{ 2 } \cdot \ p_{ 3 } = 0.7 \cdot \ 0.8 \cdot \ 0.9 = \dfrac{ 63 }{ 125 } = 0.504 \ \\ \ \\ b = a+b_{ 1 }+b_{ 2 } = 0.092+0.398+0.504 = \dfrac{ 497 }{ 500 } = 0.994$

$c = b_{ 1 }+b_{ 2 } = 0.398+0.504 = \dfrac{ 451 }{ 500 } = 0.902$

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