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fire four shots

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fire four shots

by sanju09 » Fri Mar 19, 2010 5:31 am
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) 0.036
(B) 0.084
(C) 0.36
(D) 0.64
(E) 0.964
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by harshavardhanc » Fri Mar 19, 2010 5:37 am
sanju09 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) 0.036
(B) 0.084
(C) 0.36
(D) 0.64
(E) 0.964
I used the complement method here.

prob of guns not hitting the target = .3, .4, .5 ,and .6

Prob that none of the guns hit = .3 * .4 * .5 * .6 = .036

probability that at least one hits = 1- .036 = .084

Am I correct? :(
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by sanju09 » Fri Mar 19, 2010 5:52 am
harshavardhanc wrote:
sanju09 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) 0.036
(B) 0.084
(C) 0.36
(D) 0.64
(E) 0.964
I used the complement method here.

prob of guns not hitting the target = .3, .4, .5 ,and .6

Prob that none of the guns hit = .3 * .4 * .5 * .6 = .036

probability that at least one hits = 1- .036 = .084

Am I correct? :(
not at subtraction at least, I must say
The mind is everything. What you think you become. -Lord Buddha



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by kstv » Fri Mar 19, 2010 6:03 am
Prob of getting a hit
1st shot = 0.7
2nd shot = 0.3 * 0.6=0.8 ( Prob of 1st a miss is 1-0.7)
3rd shot = 0.3*0.4*0.5 = 0.06
4th shot = 0.3*0.4*0.5*0.4= .024
Tot 0.964
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by sanju09 » Fri Mar 19, 2010 6:11 am
kstv wrote:Prob of getting a hit
1st shot = 0.7
2nd shot = 0.3 * 0.6=0.8 ( Prob of 1st a miss is 1-0.7)
3rd shot = 0.3*0.4*0.5 = 0.06
4th shot = 0.3*0.4*0.5*0.4= .024
Tot 0.964
What are we supposed to conclude now?

Is it 1 - 0.024 = 0.964, according to you, or what is your real answer?
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by harshavardhanc » Fri Mar 19, 2010 6:13 am
sanju09 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?
sanju,


The first sentence conveys that all the four shots are like a burst, i.e all at a time.
An anti aircraft gun can fire four shots at a time
however, the second sentence says that :
the first, second, third and the last shot hitting the enemy
the shots are fired one after the other. There is a difference, which can give rise to two different answers.

Can you clarify or make the question less ambiguous ?
Regards,
Harsha
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by sanju09 » Fri Mar 19, 2010 6:38 am
harshavardhanc wrote:
sanju09 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?
sanju,


The first sentence conveys that all the four shots are like a burst, i.e all at a time.
An anti aircraft gun can fire four shots at a time
however, the second sentence says that :
the first, second, third and the last shot hitting the enemy
the shots are fired one after the other. There is a difference, which can give rise to two different answers.

Can you clarify or make the question less ambiguous ?
You raised a very good point here. Imagine an anti aircraft from a very good distance if you could. Take case 1 as if the four shots were fired from the single muzzled gun, it would be bound to come out one after the other from the muzzle, so we can number it, fine. Now, take case 2 as if the four shots came out simultaneously from the anti aircraft, in this case it cannot come out from the single muzzle, it would need four different muzzles, directed slightly differently, oh DIFFERENT!! We can still number those, so the answer remains.

Do you know MUZZLE?

Now correct your subtraction work to get to the OA for this question, rest is fine.
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by harshavardhanc » Fri Mar 19, 2010 7:39 am
sanju09 wrote: correct your subtraction work to get to the OA for this question, rest is fine.
Oooops!!! what a careless mistake!!!!! shame on me!!! :(


.964 :)
Last edited by harshavardhanc on Fri Mar 19, 2010 8:33 am, edited 2 times in total.
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Harsha
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by RJ43 » Fri Mar 19, 2010 8:18 am
Assuming we only need 1 shot to land for it to go down:

Probability all shots miss = .3 * .4 * .5 *.6 = .0360, or 3.6%

Therefore, the probability of at least 1 shot landing is 1-.036 = 0.964 E.

Yes?
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by hooliganpete » Fri Mar 19, 2010 2:04 pm
Assuming we only need 1 shot to land for it to go down:
Given how detailed we have to be to answer many questions, I would hope a question like this one would specify how many shots are needed to "bring an aircraft down".
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