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An anti aircraft gun can fire four shots at a time. If the probabiliti
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18 Oct 2021, 05:33
18 Oct 2021, 05:33
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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?
(A) 0.084
(B) 0.916
(C) 0.036
(D) 0.964
(E) 0.92
(A) 0.084
(B) 0.916
(C) 0.036
(D) 0.964
(E) 0.92
Re: An anti aircraft gun can fire four shots at a time. If the probabiliti
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18 Oct 2021, 06:17
18 Oct 2021, 06:17
In the first shot, the probability of not hitting the aircraft is 1 - 0.7 = 0.3.
Similar to all other shots, the probability of all four shots missed is 0.3 x 0.4 x 0.5 x 0.6 = 0.036
=> Then the probability of bringing the aircraft down is 1 - 0.036 = 0.964
Choice D
Similar to all other shots, the probability of all four shots missed is 0.3 x 0.4 x 0.5 x 0.6 = 0.036
=> Then the probability of bringing the aircraft down is 1 - 0.036 = 0.964
Choice D
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Re: An anti aircraft gun can fire four shots at a time. If the probabiliti
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18 Oct 2021, 06:44
18 Oct 2021, 06:44
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Carcass wrote:
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?
(A) 0.084
(B) 0.916
(C) 0.036
(D) 0.964
(E) 0.92
(A) 0.084
(B) 0.916
(C) 0.036
(D) 0.964
(E) 0.92
So, we want P(at least one shot hits the aircraft)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 shot hits plane) = 1 - P(zero shots hit plane)
P(zero shots hit plane)
P(zero shots hit plane) = P(1st shot misses AND 2nd shot misses AND 3rd shot misses AND 4th shot misses)
= P(1st shot misses) x P(2nd shot misses) x P(3rd shot misses) x P(4th shot misses)
= 0.3 x 0.4 x 0.5 x 0.6
= 3/10 x 4/10 x 5/10 x 6/10
= 360/10000
= 0.036
So.......
P(at least 1 shot hits plane) = 1 - P(zero shots hit plane)
= 1 - 0.036
= 0.964
Answer: D
