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:

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

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

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

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