Algebra> An anti -aircraft gun can take a maximum ...

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?

Hrishant Goswami , 10 Years ago

Grade 10

1 Answers

Jitender Pal

Last Activity: 10 Years ago

Hello Student,

Please find the answer to your question

(a). Let us define the events as:

E₁ ≡ First shot hits the target plane,

E₂ ≡ Second shot hits the target plane,

E₃ ≡ third shot hits the target plane,

E₄ ≡ fourth shot hits the target plane

Then ATQ, P (E₁) = 0.4; P (E₂) = 0.3;

P (E₃) = 0.2; P (E₄) = 0.1

⇒ P ( $\rightarro\underset{E}{\rightarrow}$ ₁) = 1 – 0.4 = 0.6; P ( $\rightarro\underset{E}{\rightarrow}$ ₂) = 1 – 0.3 = 0.7

P ( $\rightarro\underset{E}{\rightarrow}$ ₃) = 1 – 0.2 = 0.8; P ( $\rightarro\underset{E}{\rightarrow}$ ₄) = 1 – 0.1 = 0.9

(where $\rightarro\underset{E}{\rightarrow}$ ₁ denotes not happening of E₁)

Now the gun hits the plane if at least one of the four shots hit the plane.

Also, P (at least one shot hits the plane ).

= 1 - P (none of the shots hits the plane)

= 1 - P ( $\rightarro\underset{E}{\rightarrow}$ ₁ ∩ $\rightarro\underset{E}{\rightarrow}$ ₂ ∩ $\rightarro\underset{E}{\rightarrow}$ ₃ ∩ $\rightarro\underset{E}{\rightarrow}$ ₄)

= 1 - P ( $\rightarro\underset{E}{\rightarrow}$ ₁). P ( $\rightarro\underset{E}{\rightarrow}$ ₂). P( $\rightarro\underset{E}{\rightarrow}$ ₃). P( $\rightarro\underset{E}{\rightarrow}$ ₄)

[Using multiplication thm for independent events]

= 1 - 0.6 x 0.7 x 0.8 x 0.9 = 1 - 0.3024 = 0.6976

Thanks
Jitender Pal
askIITians Faculty