# 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 the fourth shot is 0.4, 0.3, 0.2, and 0.1 respectively. What is the probability that the plane gets hit?

$

(a){\text{ 0}}{\text{.6976}} \\

(b){\text{ 0}}{\text{.7976}} \\

(c){\text{ 0}}{\text{.3024}} \\

(d){\text{ None of these}} \\

$

Answer

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Hint – In this various probabilities of hitting a plane at different rounds by an anti-craft gun is given to us. We need to find the probability of hitting a plane. So if a plane is to be hit only that it can be hit in the first round, this means it shouldn’t be hit in the second, third or fourth round. This similar concept can be extended to hit the plane in round. Use this concept along with the probability identity that probability of occurring + probability of non-occurring is equal to 1.

“Complete step-by-step answer:”

Given data

Probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1.

Then we have to find out the probability that the plane gets hit.

Probability of hitting the plane at first time $\left( {{p_1}} \right) = 0.4$

Therefore probability of not hitting the plane at first time $\left( {{q_1}} \right) = 1 - 0.4 = 0.6$

Probability of hitting the plane at second time $\left( {{p_2}} \right) = 0.3$

Therefore probability of not hitting the plane at second time $\left( {{q_2}} \right) = 1 - 0.3 = 0.7$

Probability of hitting the plane at third time $\left( {{p_3}} \right) = 0.2$

Therefore probability of not hitting the plane at third time $\left( {{q_3}} \right) = 1 - 0.2 = 0.8$

Probability of hitting the plane at fourth time $\left( {{p_4}} \right) = 0.1$

Therefore probability of not hitting the plane at fourth time $\left( {{q_4}} \right) = 1 - 0.1 = 0.9$

So the required probability (P) of hitting the plane is the addition of plane get hit first time, plane not hit first time multiplied by plane get hit second time, plane not hit first time multiplied by plane not hit second time multiplied by plane get hit third time and plane not hit first time multiplied by plane not hit second time multiplied by plane not hit third time multiplied by plane get hit fourth time.

$ \Rightarrow p = {p_1} + {q_1}{p_2} + {q_1}{q_2}{p_3} + {q_1}{q_2}{q_3}{p_4}$

$ \Rightarrow p = 0.4 + \left( {0.6 \times 0.3} \right) + \left( {0.6 \times 0.7 \times 0.2} \right) + \left( {0.6 \times 0.7 \times 0.8 \times 0.1} \right)$

$ \Rightarrow p = 0.4 + 0.18 + 0.084 + 0.0336$

$ \Rightarrow p = 0.6976$.

So this is the required probability of getting the plane hit.

Hence option (a) is correct.

Note – Whenever we face such types of problems the key concept is simply to use the property of probability that total probability is always equal to one. Try to think of each and every possible scenario in which the plane can be shot by the machine-gun once only. This will help you get on track to reach the answer.

“Complete step-by-step answer:”

Given data

Probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1.

Then we have to find out the probability that the plane gets hit.

Probability of hitting the plane at first time $\left( {{p_1}} \right) = 0.4$

Therefore probability of not hitting the plane at first time $\left( {{q_1}} \right) = 1 - 0.4 = 0.6$

Probability of hitting the plane at second time $\left( {{p_2}} \right) = 0.3$

Therefore probability of not hitting the plane at second time $\left( {{q_2}} \right) = 1 - 0.3 = 0.7$

Probability of hitting the plane at third time $\left( {{p_3}} \right) = 0.2$

Therefore probability of not hitting the plane at third time $\left( {{q_3}} \right) = 1 - 0.2 = 0.8$

Probability of hitting the plane at fourth time $\left( {{p_4}} \right) = 0.1$

Therefore probability of not hitting the plane at fourth time $\left( {{q_4}} \right) = 1 - 0.1 = 0.9$

So the required probability (P) of hitting the plane is the addition of plane get hit first time, plane not hit first time multiplied by plane get hit second time, plane not hit first time multiplied by plane not hit second time multiplied by plane get hit third time and plane not hit first time multiplied by plane not hit second time multiplied by plane not hit third time multiplied by plane get hit fourth time.

$ \Rightarrow p = {p_1} + {q_1}{p_2} + {q_1}{q_2}{p_3} + {q_1}{q_2}{q_3}{p_4}$

$ \Rightarrow p = 0.4 + \left( {0.6 \times 0.3} \right) + \left( {0.6 \times 0.7 \times 0.2} \right) + \left( {0.6 \times 0.7 \times 0.8 \times 0.1} \right)$

$ \Rightarrow p = 0.4 + 0.18 + 0.084 + 0.0336$

$ \Rightarrow p = 0.6976$.

So this is the required probability of getting the plane hit.

Hence option (a) is correct.

Note – Whenever we face such types of problems the key concept is simply to use the property of probability that total probability is always equal to one. Try to think of each and every possible scenario in which the plane can be shot by the machine-gun once only. This will help you get on track to reach the answer.

Last updated date: 17th Sep 2023

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