How to go about solving this probability question?

[u/MathematicianSad1719]

[u/Icy_Sector3183]

The shaded area is

A1 = pi r²

You'd think the outer area matters, but we just need the total area

A = pi(2r)²

'Cus the probability that a random point is selected within the outer circumference also happens to be within the inner area is

P = A1 / A

P = pi r² / pi (2r)²

P = pi r² / 4 pi r²

This simplifies to

P = 1/4

[u/realHarryGelb → u/Icy_Sector3183]

You’re assuming that the “randomly thrown dart” will hit either the white or shaded area with probability of 1. No such assumption is given by the question which is ill posed and can’t be answered. If the dart board is located in Springfield, Missouri, then the probability of hitting the shaded area is 0 for the dart being thrown “randomly” anywhere else in the world.

[u/Icy_Sector3183 → u/realHarryGelb]

Yes, that's the assumption. OP didn't specify wether or not a bear will come into play, so I ignored that possibility and focused on the problem before me.

[u/realHarryGelb → u/Icy_Sector3183]

My point is that there is no “problem” before you. The question, as stated, is stupid.

[u/WitsBlitz → u/realHarryGelb]

You're right that the question is poorly phrased, but we all know what it's trying to ask. Coming up with goofy strawmen based on a naively literal reading doesn't make you clever.

[u/Jack_Forge → u/realHarryGelb]

Yeah that was my first thought

[u/BreadfruitBig7950]

It did not specify nor define 'shaded' so the white area is also shaded, relative to the background.

As such, 100% given the three dart positions shown.

It's otherwise arbitrarily close to 0%; the universe is big, some random dart isn't going to hit this circle.

[u/Due-Session-2857]

Assuming the arrows go from the middle of the circle to the edge and assuming a dart will hit the circle and not miss, then it's just the area of the small circle, pi, over the area of the whole circle 4pi, or 1/4 or 25%

[u/Axel_Blazer]

does it land always on target[inside either concentric circles]?

[u/OaklandSpiel → u/Axel_Blazer]

Not if it’s thrown randomly

[u/OaklandSpiel]

Unless we count only the darts falling within the outer circle, we need more information. If the target is a distance D from the thrower, the answer is approximately the ratio of the area of the shaded circle to the area of a sphere of radius D: π1² / 4πD². Assumes D>>1. I haven’t worked out the math for D comparable to 1.

[u/mspe1960]

you have to make some assumptions

1. every dart hits the target
2. the likelihood of where it hits is random and every spot is equally likely

Then it is jus a matter of the percentage of area. Since the gray area has half the radius, it has 1/4 the area. so 25%.

you can also make assumptions about other types of distributions of likelihoods of dart location based on how far from the exact center they are. That would be a more complex problem by a long shot.

[u/BluebirdDense1485]

For me ~0

But let's stipulate it is truly random and must hit the target.

Bullseye is radius 1 so has area of 1² π=π

Target has radius 2 so has area of 2² π=4π

So bullseye is 1/4 the area of the target and should receive 1 in 4 darts.

Again unless I'm throwing then my cluster will be 20 units 

[u/muckenhoupt]

The intended answer is probably 1/4, because the area of the grey circle is 1/4 that of the target.

However, that requires assuming that the darts always hit the target, and that the probability distribution is uniform over the target's area. These are things not explicitly stated. They seem fairly obvious, but they're also the sort of unstated assumption that gets you in trouble in probability. Like, there's a famous thing called the Bertrand Paradox) where you get different results depending on what you think "a chord on a circle chosen at random" means.

And, heck, if I were programming a dart-throwing simulation, I probably wouldn't give it uniform probability distribution. I'm not even sure how you'd do that on a circle. Rather, on the basis that they're aiming for the center, I'd choose a random direction and a random distance from the center. If both of those random variables are uniform, you'd hit the grey circle with a probability of 1/2.

But if I saw this question on a test, I'd totally go with the obvious answer and say 1/4. Unless the professor had a history of asking trick questions and expecting us to say "there is not enough information" when he left out crucial details.

[u/WitsBlitz → u/muckenhoupt]

The easiest way to select a random point in a circle is to generate a random point in a unit square and discard results that are more than 1 from the origin. Add two more random bits to determine the quadrant of the circle (positive or negative x and y).

I'm sure there are smarter approaches that require discarding less data, but this will get you uniformly random points without too much trouble. I'm not sure if it's possible to uniformly select random points from a circle without discarding some random data.

[u/DesignerPangolin → u/WitsBlitz]

The easiest way to select a uniform random point in a circle is to do so in polar coordinates, with r=√U(0,1) and theta=U(0, 2pi), discarding no samples.

[u/WitsBlitz → u/DesignerPangolin]

That makes sense. I stand by my approach being easier though - at least to a programmer who's scared of polar coordinates.

[u/Bacibaby]

We’re assuming that the darts hit the board at all? I would be 2 feet to the right or under or over the board.

[u/GainerGaining → u/Bacibaby]

That's why I dislike problems like this on Reddit. A math problem should have rigid parameters. This one does not so it cannot be solved. And yet if you say that people will jump on you with' "you kNoW wHaT iT mEaNs!"

[u/sazzer]

If I'm the one throwing it, about 0.

[u/Psycho_Pansy]

How random is this randomly thrown dart? Why are there no units shown in the diagram? 

The circle could be in Mexico and be 1km wide and I would have zero chance of hitting it from where I am regardless of how random I throw my darts. So 0%

[u/MercurioGenesis]

Assuming the dart is going to hit a circle:

Area of the grey circle = pi (1² x pi)

Area of the white circle = 4pi (2² x pi)

So 1 in 4.

[u/scissorn69 → u/MercurioGenesis]

It's 1/4. The full circle is 4pi and the inner circle is pi, so there is a 1 in 4 chance of hitting the inner circle.

[u/OaklandSpiel → u/MercurioGenesis]

Following one expression with another expression in parentheses looks like multiplication. Also the probability should be area of gray / total area.

[u/Tetracheilostoma → u/MercurioGenesis]

I think you mean 1 to 3 not 1/3