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

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.21

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6976

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 0.6976
Let $p_{1} = 0.4, p_{2} = 0.3, p_{3} = 0.2$ and $p_{4} = 0.1$
P(the sun hits the plane) = P(the plane is hit in once)
= 1 – P (the plane is hit in none of the shots)
$= 1 - (1 - p_{1}) (1 - p_{2}) (1 - p_{3}) (1 - p_{4}) = 0.6976$

Suggest Corrections

Similar questions

Q. An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it.The probabilities of hitting the plane at the first, second, third and fourth shot are

0.4, 0.3, 0.2

and

0.1

respectively. What is the probability that the gun hits the plane?

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

[CEE 1993; IIT Screening]

Karan can take a maximum of four shots at an enemy moving away from him. The probability of hitting the enemy at the first, second, third and fourth shots are 0.3, 0.4, 0.2 and 0.1 respectively. What is the probability that the enemy is hit when all the four shots are fired?

A gun can fire four shots at a time. If the probabilities of the first, second, third and the last shot hitting the enemy are 0.7, 0.6, 0.5 and 0.4, what is the probability that four shots aimed at an enemy will hit the target?

Join BYJU'S Learning Program

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

The correct option is D 0.6976Let p1=0.4,p2=0.3,p3=0.2 and p4=0.1 P(the sun hits the plane) = P(the plane is hit in once) = 1 – P (the plane is hit in none of the shots) =1–(1–p1)(1–p2)(1–p3)(1–p4)=0.6976

The correct option is D 0.6976
Let $p_{1} = 0.4, p_{2} = 0.3, p_{3} = 0.2$ and $p_{4} = 0.1$
P(the sun hits the plane) = P(the plane is hit in once)
= 1 – P (the plane is hit in none of the shots)
$= 1 - (1 - p_{1}) (1 - p_{2}) (1 - p_{3}) (1 - p_{4}) = 0.6976$