NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

An anti-aircraft gun can take a maximum of four slots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth slots 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. Calculate the probability of missing at each slot:

Let $P(H_i)$ be the probability of hitting at slot $i$ , and $P(M_i)$ be the probability of missing at slot $i$ .

$P(M_1) = 1 - 0.4 = 0.6$
$P(M_2) = 1 - 0.3 = 0.7$
$P(M_3) = 1 - 0.2 = 0.8$
$P(M_4) = 1 - 0.1 = 0.9$

2. Calculate the probability of missing in all four slots:

Since the shots are independent events, we multiply the individual missing probabilities:

P(\text{Miss all}) = P(M_1) \times P(M_2) \times P(M_3) \times P(M_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

3. Calculate the probability of hitting the plane:

The probability of hitting the plane is at least one hit occurring:

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

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

P(\text{Hit}) = 0.6976

The correct option is (c) 0.6976.