probability question?

Question

What is the probability that three points selected at random in a sphere will be in the same hemisphere.Please explain the procedure also

Answer 1

1/4.
The first one you choose can come from either hemisphere,we are not concerned with it because it has nothing to do with the probability.
as there are two hemispheres, the probability that the second one come from the same hemisphere is 1/2.
similarly the probability of getting the third point from the same hemisphere as the first one is 1/2
required probability=
P(Ist pt comes from a hemisphere(intersection)2nd pt comes from the same hemisphere (intersection) 3rd pt comes from the same hemisphere)=P(Ist pt comes from a hemisphere)*P(2nd pt comes from the same hemisphere)*P(3nd pt comes from the same hemisphere)
{because choice of pts are independent events}
=1/2*1/2=1/4
Alternative way:-
The possible ways in which you can choose those pts are as follows:-
1st & 2nd are of same hemisphere 3rd is not
1st & 3nd are of same hemisphere 2rd is not
3rd & 2nd are of same hemisphere 1st is not
all of the same hemisphere
Since there are 4 possible outcomes & only 1 favourable to our desired event
The required probability=1/4

Answer 2

Revision: 100%

Let a point land randomly on the sphere. Rotate the sphere so that this point is at the North Pole. Now start looking at the possible great circles that contain this point. These are the longitudinal lines. Every point on the globe has a longitudinal coordinate. Choose the great circle that has the same longitude as the second point. Therefore both points will be on the same great circle. Now rotate the sphere so this great circle is labeled with 0º longitude, and let the third point fall on the sphere. Either it lands in the Eastern or Western hemisphere. Either way it is in the same hemisphere as the other two.

Original:
50%

Given 1 point, draw a great circle that contains that point. Now allow another point to be placed on the sphere at random. This point is either in the northern or southern hemisphere. No matter where it is, it is in the same hemisphere as the first point. Now choose that to be your hemisphere.

Given a third point, there will be a 50% chance that it is in the chosen hemisphere. Therefore the over all probability is 50%.

Answer 3

The question is ambiguous...is the hemisphere (1)pre-determined, or can you construct a new hemishpere after each point placement(2)?
If (1) correct answer is 25%
If (2) correct answer is 50%
Both possibilities are adequately discussed above.

Better questions would be:
In the case of (1)...If three points are randomly selected on the earth's surface, what is the probability that all three will be either above the equator or below it?
In the case of (2)...If three points have been randomly placed on a sphere, what is the probability that a hemisphere can be constructed to enclose all three?

Answer 4

I would say 1/4 . The first one you pick is irrelevant to the probability. The second one has a 1/2 probability of being in the same hemisphere as the first and the third one has a probability of 1/2 of being in the same hemisphere as the other two, 1/2 * 1/2 = 1/4

Answer 5

Perhaps 1/3rd...because there is a 2/3 chance of them being in different hemispheres..

Answer 6

umm spin the globe stop it at random three times and see what ypu come up with but i say 50 percent

Answer 7

there can be infenite or count less points on a circel
so one of the data is infinet
ans is also infines
as to get probebelity you have to divied a number by infinity
wich is infinity again

Answer 8

100%

Answer 9

hemi...1/2

so .5*.5*.5= .125