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Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.4 Question 18 Maths Textbook Solution.

Answers (1)

Answer:0.6976

Hint:P\left ( A \right )+P\left ( \bar{A} \right )=1

Given: 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 first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively.

Solution:

Let,

      P (Gun hits the plane) = 1 – (Gun does not hit the plane)

       \begin{aligned} &P(A)=1-P(\bar{A}) \\ &\begin{aligned} P(\bar{A}) &=(1-0.4)(1-0.3)(1-0.2)(1-0.1) \\ &=0.6 \times 0.7 \times 0.8 \times 0.9 \\ &=0.3024 \\ P(A) &=1-0.3024 \\ P(A) &=0.6976 \end{aligned} \end{aligned}

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