fire four shots

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?

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

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

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?

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 ?

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.

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

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?

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".