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:

$$\begin{gathered} p_1=0.7 \\ {p}_2=0.8 \\ {p}_3=0.9 \\ {n}_1=1-p_1=1-0.7={\displaystyle \frac{3}{10}}=0.3 \\ {n}_2=1-p_2=1-0.8={\displaystyle \frac{1}{5}}=0.2 \\ {n}_3=1-p_3=1-0.9={\displaystyle \frac{1}{10}}=0.1 \\ a=p_1\cdot n_2\cdot n_3+n_1\cdot p_2\cdot n_3+n_1\cdot n_2\cdot p_3=0.7\cdot 0.2\cdot 0.1+0.3\cdot 0.8\cdot 0.1+0.3\cdot 0.2\cdot 0.9={\displaystyle \frac{23}{250}}=0.092 \end{gathered}$$

$$\begin{gathered} b_1=p_1\cdot p_2\cdot n_3+n_1\cdot p_2\cdot p_3+p_1\cdot n_2\cdot p_3=0.7\cdot 0.8\cdot 0.1+0.3\cdot 0.8\cdot 0.9+0.7\cdot 0.2\cdot 0.9={\displaystyle \frac{199}{500}}=0.398 \\ {b}_2=p_1\cdot p_2\cdot p_3=0.7\cdot 0.8\cdot 0.9={\displaystyle \frac{63}{125}}=0.504 \\ b=a+b_1+b_2=0.092+0.398+0.504={\displaystyle \frac{497}{500}}=0.994 \end{gathered}$$

$c=b_1+b_2=0.398+0.504={\displaystyle \frac{451}{500}}=0.902$

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