Three shooters

Three shooters each fire once at the same target. The first hits the target with a probability of 0.7, the second with 0.8, and the third with 0.9. What is the probability of hitting the target:

a) exactly once?
b) at least once?
c) at least twice?

Final 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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