Question

An anti-aircraft gun takes 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:

• A. 0.2412
• B. 0.21
• C. 0.16
• D. 0.6976

Answer 1

Answer: (D)

0.6976

Answer 2

The probability of hitting the gun in the first shot i.e. $p_1 = 0.4$ Similarly, $p_2 = 0.3, p_3 = 0.2, p_4 = 0.1$ The probability of not hitting the plane in their first, second,.......fourth shot will be $q_1 = 0.6, q_2 = 0.7, q_3 = 0.8, q_4 = 0.9$ Now, the probability of not hitting the plane will be $q = q_1 \times q_2 \times q_3 \times q_4$ $= 0.6 \times 0.7 \times 0.8 \times 0.9 = 0.3044$ Now the probability that the gun hits the plane is p = 1 - q i.e. p = 1 - 0.3044 i.e. p = 0.6976