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25 Oct 2016, 08:42
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:01) correct
37%
(01:42)
wrong
based on 270
sessions
History
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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
Source: https://www.GMATinsight.com
(A) 0.084
(B) 0.916
(C) 0.036
(D) 0.964
(E) 0.92
Source: https://www.GMATinsight.com
Updated on: 02 Apr 2020, 08:05
Originally posted by BrentGMATPrepNow on 25 Oct 2016, 09:17.
Last edited by BrentGMATPrepNow on 02 Apr 2020, 08:05, edited 1 time in total.
Last edited by BrentGMATPrepNow on 02 Apr 2020, 08:05, edited 1 time in total.
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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:
General Discussion
25 Oct 2016, 09:18
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Sum of probabilities of each shot= 0.7+0.3*0.6+0.3*0.4*0.5+0.3*0.4*0.5*0.4
=0.964
IMO D
Sent from my iPhone using GMAT Club Forum mobile app
=0.964
IMO D
Sent from my iPhone using GMAT Club Forum mobile app
