Three shooters

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

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

Correct answer:

a =  0.092
b =  0.994
c =  0.902

Step-by-step explanation:

$$\begin{gathered} p_1=0.7 \\ {p}_2=0.8 \\ {p}_3=0.9 \\ {n}_1=1-p_1=1-0.7=\frac{3}{10}=0.3 \\ {n}_2=1-p_2=1-0.8=\frac{1}{5}=0.2 \\ {n}_3=1-p_3=1-0.9=\frac{1}{10}=0.1 \\ a=p_1\cdot n_2\cdot n_3+n_1\cdot p_2\cdot n_3+n_1\cdot n_2\cdot p_3=0.7\cdot 0.2\cdot 0.1+0.3\cdot 0.8\cdot 0.1+0.3\cdot 0.2\cdot 0.9=0.092 \end{gathered}$$

$$\begin{gathered} b_1=p_1\cdot p_2\cdot n_3+n_1\cdot p_2\cdot p_3+p_1\cdot n_2\cdot p_3=0.7\cdot 0.8\cdot 0.1+0.3\cdot 0.8\cdot 0.9+0.7\cdot 0.2\cdot 0.9=\frac{199}{500}=0.398 \\ {b}_2=p_1\cdot p_2\cdot p_3=0.7\cdot 0.8\cdot 0.9=\frac{63}{125}=0.504 \\ b=a+b_1+b_2=0.092+0.398+0.504=0.994 \end{gathered}$$

$c=b_1+b_2=0.398+0.504=0.902$

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