Note: (a,b) is the open interval from a to b, a>b
1) Exactly 2: P1 lies in either (r,R) or (-R,r). P = 2(R-r)/2R
P2 must lie in the other interval. P = (R-r)/2R
Multiplying and reducing gives p(2) = (R-r)^2 / 2R^2

2) Exactly 1: P1 lies in either (r,R) or (-r,R), P = 2(R-r)/2R

P2 lies in (-r,r). P = 2r/2R

doubling since we can switch P1 and P2

Multiplying and reducing gives P(1) = 2r(R-r) / R^2

3) Exactly 0 : There are 3 cases:

a) P1 and P2 both lie in (-R,-r). P = (R-r)^2/4R^2

b) P2 and P1 both lie in (-r,r). P =( 2r/2R)^2 = r^2/R^2

c) P1 and P2 lie both on (r,R). P = (R-r)^2/4R^2

Getting a common denominator and adding gives P(0) = [2(R-r)^2 + 4r^2] / 4R^2

Thanks, Ian. Glad you liked it. Did you notice that at R=3 (when r=1) probability function P(R) reaches its maximum P(3)=2/3? And the fact that P(R) approaches 1/2 as R goes to infinity also surprises me.

It it quite obvious as you look at the function but isn't it interesting from the "probability point of view"?

Indeed. I understand why P(R/r=3)=2/3 have to be a maximum value. But how can you say it without using differentiation and P(R) function? I know there should be simple explanation. And the question about the probability approaching 1/2 at infinity is still up in the air.

I also want to suggest harder problem which I am still trying to solve. Go one dimension higher and now consider two concentric circles instead of two pairs of points. Find the probability for a random section inside the big circle to cross the smaller circle.

I'm hoping I have interpreted the 1/2 thing correctly. If not maybe you can explain again.

r is fixed, R can vary, choose your 2 points randomly and you want the section to cross both P1 and P2 which are chosen at random

1) As R----> inf then the 2 outside intervals (-R,-r) and (r,R) become so large and the inside interval (-r,r) becomes so small in comparison that the the 2 random points almost never lie in (-r,r)

So when you choose P1 it is almost guaranteed to be to be in either (-R,-r) or (r,P) making that probability ---->1

Choosing P2 requires that that it be chosen in the other outside interval which covers almost one-half of the number line making the probability ---->1/2

Since both must must happen then multiply and (1)(1/2) = 1/2

Choose 2 points P1 and P2 at random inside the large circle. What is the probability that the line joining P1 and P2 will cross the smaller circle. I will show 2 methods.

1) P1 is in the small circle. P = kr^2 / kR^2 = r^2 / R^2

P2 must lie in the ring. P = k(R^2-r^2) / kR^2 = (R^2 - r^2)/ R^2

Mulitply since both must happen

P(crossing) = r^2(R^2 - r^2) / R^4

We repeat the process for P2 in the small circle and P1 in the ring and get the same answer
so we double.

P(crossing)

= [r^2/R^2] [(R^2-r^2)/R^2] *2

= 2r^2(R^2 - r^2) / R^4

2) P(crossing)

= 1 - both lie in the ring - both lie in the small circle

I stand corrected. Thank You. That's what happens when you are in too big of a hurry. You are also correct in that there is more to this problem than meets the eye. Thanks for the update and I will soldier on - Ian

Note: (a,b) is the open interval from a to b, a>b 1)

Exactly 2: P1 lies in either (r,R) or (-R,r). P = 2(R-r)/2R P2 must lie in the other interval. P = (R-r)/2R Multiplying and reducing givesp(2) = (R-r)^2 / 2R^22)

Exactly 1: P1 lies in either (r,R) or (-r,R), P = 2(R-r)/2RP2 lies in (-r,r). P = 2r/2R

doubling since we can switch P1 and P2

Multiplying and reducing gives

P(1) = 2r(R-r) / R^23)

Exactly 0: There are 3 cases:a) P1 and P2 both lie in (-R,-r). P = (R-r)^2/4R^2

b) P2 and P1 both lie in (-r,r). P =( 2r/2R)^2 = r^2/R^2

c) P1 and P2 lie both on (r,R). P = (R-r)^2/4R^2

Getting a common denominator and adding gives

P(0) = [2(R-r)^2 + 4r^2] / 4R^2Check:

P(2) + P(1) + P(0)= (R-r)^2/2R^2 + 2r(R-r)/R^2 +[ 2(R-r)^2 + 4R^2 ] / 4R^2Getting a common denominator and simplifying gives:

4R^2/4R^2 = 1Thanks for the great problem. I enjoyed it. - Ian

Thanks, Ian. Glad you liked it. Did you notice that at R=3 (when r=1) probability function P(R) reaches its maximum P(3)=2/3? And the fact that P(R) approaches 1/2 as R goes to infinity also surprises me.

It it quite obvious as you look at the function but isn't it interesting from the "probability point of view"?

Nice analysis. Makes sense.

Consider this:

P(1) = 2r(R-r) / R^2 and keeping r constant

Differentiate P(1) wrt R and set equal to 0 yields R = 2r as a stationary point.

Differentiating again to get the second derivative and substituting R = 2r yields the constant -2 which proves R = 2r gives the maximum value of P(1).

Substituting R = 2r back into P(1) gives the maximum value of P(1) = 1/2r^2

I think this is worth a video on bprp.

Indeed. I understand why P(R/r=3)=2/3 have to be a maximum value. But how can you say it without using differentiation and P(R) function? I know there should be simple explanation. And the question about the probability approaching 1/2 at infinity is still up in the air.

I also want to suggest harder problem which I am still trying to solve. Go one dimension higher and now consider two concentric circles instead of two pairs of points. Find the probability for a random section inside the big circle to cross the smaller circle.

I'm hoping I have interpreted the 1/2 thing correctly. If not maybe you can explain again.

r is fixed, R can vary, choose your 2 points randomly and you want the section to cross both P1 and P2 which are chosen at random

1) As R----> inf then the 2 outside intervals (-R,-r) and (r,R) become so large and the inside interval (-r,r) becomes so small in comparison that the the 2 random points almost never lie in (-r,r)

So when you choose P1 it is almost guaranteed to be to be in either (-R,-r) or (r,P) making that probability ---->1

Choosing P2 requires that that it be chosen in the

other outside interval which covers almostone-halfof the number line making the probability ---->1/2Since both must must happen then multiply and (1)(1/2) = 1/2

2) P(2) = (R-r)^2/2R^2

lim P(2) as R--> inf

= lim as R -->inf [(R-r)^2/2R^2]

= lim as R --> inf [(R^2 -2rR +r^2) / 2R^2]

dividing through by R^2

= lim as R --> inf [ 1 - 2r/R + r^2/R^2] /2

= [1-0+0]/2

= 1/2

@Ian Fowler Exactly. I want the section that connects my two random picked points to include one or both of x=+-r.

I got your explanation of why it approaches infinity. Sounds reasonable, thanks!

Circle problemI shall use k for "pi"Area of large circle =

kR^2Area of small circle =

kr^2Area of ring = kR^2 - kr^2 =

k(R^2 - r^2)Choose 2 points P1 and P2 at random inside the large circle. What is the probability that the line joining P1 and P2 will cross the smaller circle. I will show 2 methods.

1) P1 is in the small circle. P = kr^2 / kR^2 =

r^2 / R^2P2 must lie in the ring. P = k(R^2-r^2) / kR^2 =

(R^2 - r^2)/ R^2Mulitply since both must happen

P(crossing) = r^2(R^2 - r^2) / R^4

We repeat the process for P2 in the small circle and P1 in the ring and get the same answer so we

double.P(crossing)

= [r^2/R^2] [(R^2-r^2)/R^2] *2

=

2r^2(R^2 - r^2) / R^42) P(crossing)

= 1 - both lie in the ring - both lie in the small circle

= 1 - [k(R^2-r^2)/kR^2][k(R^2-r^2)/kR^2] - [kr^2/kR^2][kr^2/kR^2]

cancelling out the "k"s

= 1 - [(R^2 - r^2)^2/R^4] - [r^4/R^4]

R^4 is the common denominator

= [ R^4 - (R^4 - 2r^2R^2 + r^4) - r^4 ] / R^4

= [R^4 - R^4 + 2r^2R^2 - r^4 - r^4] / R^4

= [ 2r^2R^2 - 2r^4] / R^4

factoring the numerator

=

2r^2(R^2 - r^2) / R^4BTW you can still apply your analysis when r is fixed and R ---> inf

Now the chances are that both P1 and P2 will be in the ring and P(crossing) ---> 0

I stand corrected. Thank You. That's what happens when you are in too big of a hurry. You are also correct in that there is more to this problem than meets the eye. Thanks for the update and I will soldier on - Ian

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