# An anti-aircraft gun can take a maximum of four shots to an enemy plane moving away from it. The probability of hitting the plane at first, second, third & fourth shots are ${\text{0}}{\text{.4,0}}{\text{.3,0}}{\text{.2,0}}{\text{.1}}$ respectively . What is the probability that the plane gets hit?

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Hint : Find the probability of the plane not getting hit and subtract it from total probability

Let ${H_1},{H_2},{H_3}\& {H_4}$ be the probabilities of hitting the plane at first, second, third & fourth shots.

$

P({H_1}) = 0.4 \\

P({H_2}) = 0.3 \\

P({H_3}) = 0.2 \\

P({H_4}) = 0.1 \\

$

Here we will find the probability of the plane doesn’t get hit at first , second , third & fourth shots respectively ,

$

P({\overline H _1}) = 0.6{\text{ }} \\

P({\overline H _2}) = 0.7{\text{ }} \\

P({\overline H _3}) = 0.8{\text{ }} \\

P({\overline H _4}) = 0.9 \\

$

Total probability that the plane doesn’t gets hit is :-

$P({\overline H _1})P({\overline H _2})P({\overline H _3})P({\overline H _4}) = (0.6){\text{ (}}0.7){\text{ (}}0.8){\text{ (}}0.9) = 0.3024$

Now the probability that the plane gets hit is,

${\text{1 - }}0.3024 = 0.6976.$

Answer = ${\text{0}}{\text{.6976}}$

Note :- In these types of questions of probability always try to find the probability of not happening of an event . Then subtract it with one to get the probability of happening of that event . Since it is easy to solve.

Let ${H_1},{H_2},{H_3}\& {H_4}$ be the probabilities of hitting the plane at first, second, third & fourth shots.

$

P({H_1}) = 0.4 \\

P({H_2}) = 0.3 \\

P({H_3}) = 0.2 \\

P({H_4}) = 0.1 \\

$

Here we will find the probability of the plane doesn’t get hit at first , second , third & fourth shots respectively ,

$

P({\overline H _1}) = 0.6{\text{ }} \\

P({\overline H _2}) = 0.7{\text{ }} \\

P({\overline H _3}) = 0.8{\text{ }} \\

P({\overline H _4}) = 0.9 \\

$

Total probability that the plane doesn’t gets hit is :-

$P({\overline H _1})P({\overline H _2})P({\overline H _3})P({\overline H _4}) = (0.6){\text{ (}}0.7){\text{ (}}0.8){\text{ (}}0.9) = 0.3024$

Now the probability that the plane gets hit is,

${\text{1 - }}0.3024 = 0.6976.$

Answer = ${\text{0}}{\text{.6976}}$

Note :- In these types of questions of probability always try to find the probability of not happening of an event . Then subtract it with one to get the probability of happening of that event . Since it is easy to solve.

Last updated date: 20th Sep 2023

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