anuu wrote:A circular target of unit radius is divided into four annular zones with outer radii 1/4, 1/2, 3/4, and 1, respectively. Suppose ten shots are fired independently and at random into the target.
a) What is the prob. that at most three shots land in the zone bounded by the circles of radius 1/2 and 1.
b) If five shots land inside the disk of radius 1/2, find the probability that at least one is in the disk of radius 1/4.
manpsingh's explanation above is excellent. These questions are nothing at all like real GMAT questions, however. For one thing, you'd never need to know what 'annular zones' are on the GMAT, and for another, the computation involved is too cumbersome for the test. I imagine many people will not understand even what the question is asking because of the language used. The question is describing something a bit like a dartboard: we have four concentric circles with radii 1/4, 1/2, 3/4 and 1.
What you may, on rare questions, need to understand is how to calculate geometric probabilities. If we take a much simpler setup:
A circle C of radius 1 is contained within a circle D of radius 2. If a random point is selected from within circle D, what is the probability this point is contained in circle C?
this is just a complicated way of asking for the ratio of the area of C to the area of D (the answer is (Pi*(1)^2)/(Pi*(2)^2) = 1/4). That is, geometric probability questions are really just asking for ratios of areas.
While it's not something you'll see on the GMAT, we can apply this to the second question:
If five shots land inside the disk of radius 1/2, find the probability that at least one is in the disk of radius 1/4.
Here we are asked the probability that 'at least one' is in the disk of radius 1/4. There would be several cases to consider if we did the problem directly (exactly one, exactly two, exactly three, exactly four and exactly five shots in the smaller circle). Because that looks like a lot of work, it's a good idea to reverse the problem: find the probability that NO shots are in the smaller circle, and then subtract this from 1. So the probability that at least 1 shot is in the smaller circle is equal to:
1 - Probability(0 shots are in the small circle)
Now, the probability that one shot is in the small circle is just equal to the ratio of the area of the small circle to the larger circle:
Pi*(1/4)^2 / Pi*(1/2)^2 = 1/4
So the probability the first shot is NOT in the small circle is 1 - 1/4 = 3/4. Thus the probability all five shots are NOT in the small circle is (3/4)^5, and subtracting this from 1, our answer is 1 - (3/4)^5.
