Question

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?

Answer 1

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.