Three sharp shooters shoot and hit a rapidly revolving spherical taget.?

Question

Find the probability that all three points of impact are on the same hemisphere.

Answer 1

I'm going to try my answer again.

It is impossible for any two shots to be in different hemispheres, because there is no way for the second point of impact to be more than half of the sphere away from the first point of impact, as anything more than half of the sphere away from the first point in any one direction, is the same as being less than half a sphere away from first shot in the diametrically opposed direction.

And, since the third shot would have to fall somewere in between the two most distant possible positions of the points of impact of the two first shots, it, too, would have to fall into the same hemisphere of the two most distant to each other.

So, 100%.

In order to make the probabilty any smaller than 100%, you would have to include a fourth shot.

And, no, this would not have to fall into the realm of nonastronomical mathematics, because all you would have to do is replace the sharpshooters with near-earth-object impacts, and it fits right in.

Answer 2

The first two hits are always in the same hemisphere as each other -- they can't be more than 1/2 the sphere apart. So it's just a question of the distribution for the third hit.

After several milliseconds of thinking it over, I expect the third hit will have a 50% probability of being in the same hemisphere with the first two.

Answer 3

Assuming that everything is random and that every shot hits the sphere.

The probability that the first shot is in the same hemisphere as the first shot is 1 (i.e., 100% -- don't laugh, this is a necessary first step if you want your analysis to be complete).
The probability that the second shot is in the same hemisphere as the first, is 0.5 (50%)
The probability that the third shot is in the same hemisphere as the first, is 0.5 (50%).

For the described situation to occur, all three events must occur. Therefore, we multiply the probabilities:
1*0.5*0.5 = 0.25 (= 25% = 1 in 4)

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you do realise that these particular sharpshooters belong in mathematics rather than astronomy, right?

Answer 4

I agree with previous analysis in that the first 2 shots must be in the same hemisphere. However, the probability for the 3rd shot depends on the relative positions of the first 2. I think the problem abstracts to a circle with 3 points positioned randomly on the circumference. What is the probabiliy that all 3 fall within a semicircle? Its beyond me!.

Answer 5

For the first shot the chances of hitting the sphere is 50/50.

The chances of the second shot hitting sphere is 50/50 and the chances of hitting the same hemisphere as the first shot is 50/50. Therefore the chances of the second shot being in the same hemisphere has first shot 25%.

The chances of the third shot hitting the sphere, is the same as the first and second shot, 50/50. Again the chances of hitting the same hemisphere as the first and second shot, again is 50/50. Therefore the chances of the third shot being in the same hemisphere as the first and second shots is 25%.

Therefore, the chances of all three shots in the same hemisphere is = .5 X .25 X .25 = 3.125%. This assumes that the sphere is not rotating or in any other motion.

Answer 6

Any three points on a sphere lie in the same hemisphere, with the hemisphere orientation suitably defined. At worst, two points would be antipodal, and the third would then define the great circle that divided the hemispheres.

Answer 7

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Answer 8

probably not.