An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are $0.4,0.3,0.2$ and $0.1$ respectively. What is the probability that at least one shot hits the plane?

Home » An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are and respectively. What is the probability that at least one shot hits the plane?

An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are $0.4,0.3,0.2$ and $0.1$ respectively. What is the probability that at least one shot hits the plane?

Given:Let $A, B, C$ and $D b e$ first second third and fourth shots whose probability of hitting the plane is given i.e, $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3, \mathrm{P}(\mathrm{C})=0.2$ and $\mathrm{P}(\mathrm{D})=0.1$ respectively
$\Rightarrow \mathrm{P}(\bar{A})=0.6$ and $\mathrm{P}(\bar{B})=0.7$ and $\mathrm{P}(\bar{C})=0.8$ and $\mathrm{P}(\bar{D})=0.9$
Here, $\mathrm{P}$ (atleast one shot hits the plane) $=1-\mathrm{P}$ (none of the shots hit the plane)
$\begin{array}{l} =1-\mathrm{P}(\bar{A} \cap \bar{B} \cap \bar{C} \cap \bar{D}) \\ =1-[\mathrm{P}(\bar{A}) \times \mathrm{P}(\bar{B}) \times \mathrm{P}(\bar{C}) \times \mathrm{P}(\bar{D})] \\ =1-[0.6 \times 0.7 \times 0.8 \times 0.9] \\ =1-0.3024 \\ =0.6976 \end{array}$
Therefore, The probability that atleast one shot hits the plane is $0.6976 .$