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Doctor Q! (Math peoples please come)
- Thread starter fireshot91
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The first problem is kind of silly. I don't think there's enough information. I guess you could assume a uniform distribution inside of the circle and zero probability outside of the circle. If the dart has yet to be thrown, and you assume the person throwing it is aiming for the circle (presumably the center in order to minimize the chances of the dart landing outside), I find it hard to believe a uniform distribution is possible.
I might like the problem better if it said "given that a dart has landed in this circle, what is the probability it landed outside of the triangle?" That way you don't have to assume the thrower was aiming at anything in particular.
You could also assume the circle is the entire universe and the throw is completely random. I'd imagine the circle as the ground with walls extending upward. You throw something up in the air. It might bounce off the walls a few times, but it eventually has to land somewhere on the circle. But why use the image of a dart throw in that case? Overall, a poorly worded problem.
Or you could just cut the BS and ask "what percentage of this circle is shaded?"
The second problem doesn't require any trigonometry. Plane geometry suffices.
I might like the problem better if it said "given that a dart has landed in this circle, what is the probability it landed outside of the triangle?" That way you don't have to assume the thrower was aiming at anything in particular.
You could also assume the circle is the entire universe and the throw is completely random. I'd imagine the circle as the ground with walls extending upward. You throw something up in the air. It might bounce off the walls a few times, but it eventually has to land somewhere on the circle. But why use the image of a dart throw in that case? Overall, a poorly worded problem.
Or you could just cut the BS and ask "what percentage of this circle is shaded?"
The second problem doesn't require any trigonometry. Plane geometry suffices.
swiftaw
macrumors 603
For the first one, assuming each point is equally likely for the dart to land, then the probability of landing in the shaded area = area of shaded part / total area
total area = 16*pi
area of shaded part = 16*pi - 12sqrt(3)
so probability = 1 - (3*sqrt(3) / 4*pi) = 0.5865
Where did these come from. Well, total area is easy, it's a circle so area is pi*r^2 = pi*4^2 = 16*pi
area of shaded part = area of circle - area of triangle.
Okay, how to find the area of the triangle. Divide up the big triangle into 3 identical isosceles triangles by drawing lines from the center to each vertex. Each of those isosceles triangles has two sides of length 4 and an angle of 120 degrees. Basic trig shows the area of each is 4*sqrt(3), so the big triangle has area 12*sqrt(3).
total area = 16*pi
area of shaded part = 16*pi - 12sqrt(3)
so probability = 1 - (3*sqrt(3) / 4*pi) = 0.5865
Where did these come from. Well, total area is easy, it's a circle so area is pi*r^2 = pi*4^2 = 16*pi
area of shaded part = area of circle - area of triangle.
Okay, how to find the area of the triangle. Divide up the big triangle into 3 identical isosceles triangles by drawing lines from the center to each vertex. Each of those isosceles triangles has two sides of length 4 and an angle of 120 degrees. Basic trig shows the area of each is 4*sqrt(3), so the big triangle has area 12*sqrt(3).
It is a parallelogram.
The little triangles mean that the lines are parallel.
Basically ----->---- is parallel to ----->----------
and ------>-->----- is parallel to -------->->------
At least, that's the notation we used...
sushi
Moderator emeritus
Yes, I know that it is. If you read my post entirely you will see that I come to that conclusion.
Note, I may have not expressed my point well and for that I apologize.
Anyhow, an old adage for engineers, software developers and pilots, never assume! Always verify! 🙂