Probability Answers

Probability - Quant/Math - CAT 2008

Question 4 the day: July 2, 2003

The question for the day is from the topic of Probability.

An anti aircraft gun can fire four shots at a time. If the probabilities of the first, second, third and the last shot hitting the enemy aircraft are 0.7, 0.6, 0.5 and 0.4, what is the probability that four shots aimed at an enemy aircraft will bring the aircraft down?

(1)	0.084			(2)	0.916
(3)	0.036			(4)	0.964

Correct Answer - (4)

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Solution:


The enemy aircraft will be brought down even if one of the four shots hits the aircraft.

The opposite of this situation is that none of the four shots hit the aircraft.

The probability that none of the four shots hit the aircraft is given by (1-0.7)(1-0.6)(1-0.5)(1-0.4) = 0.3*0.4*0.5*0.6 = 0.036

So, the probability that at least one of the four hits the aircraft = 1 � 0.036 = 0.964.