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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 |
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(2) |
0.916 |
| (3) |
0.036 |
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(4) |
0.964 |
Correct Answer - (4)
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.
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