NIMCET 2009 — Mathematics PYQ

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Question

NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

An anti-aircraft gun takes a maximum of four 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. The probability that the gun hits the plane then is:

Choose the correct answer:

Explanation

1. Define the Probabilities of Hitting

Let $P(H_1), P(H_2), P(H_3),$ and $P(H_4)$ be the probabilities of hitting the plane on the 1st, 2nd, 3rd, and 4th shots respectively.

$P(H_1) = 0.4$
$P(H_2) = 0.3$
$P(H_3) = 0.2$
$P(H_4) = 0.1$

2. Find the Probabilities of Missing

The probability of missing a shot is $1 - P(\text{Hit})$ .

$P(\text{Miss}_1) = 1 - 0.4 = 0.6$
$P(\text{Miss}_2) = 1 - 0.3 = 0.7$
$P(\text{Miss}_3) = 1 - 0.2 = 0.8$
$P(\text{Miss}_4) = 1 - 0.1 = 0.9$

3. Calculate the Probability of Missing All Shots

The gun fails to hit the plane only if it misses all four shots. Since the shots are independent:

P(\text{Miss all}) = P(\text{Miss}_1) \times P(\text{Miss}_2) \times P(\text{Miss}_3) \times P(\text{Miss}_4)

P(\text{Miss all}) = 0.6 \times 0.7 \times 0.8 \times 0.9

P(\text{Miss all}) = 0.42 \times 0.72

P(\text{Miss all}) = 0.3024

4. Calculate the Probability of Hitting the Plane

The probability of hitting the plane is the complement of missing all shots:

P(\text{Hit}) = 1 - P(\text{Miss all})

P(\text{Hit}) = 1 - 0.3024

P(\text{Hit}) = 0.6976

Correct Option: (c) 0.6976