Single-Variable-Calculus · Unit 20 · Video 2 · Interactive Practice

One Percent Risk: Calibrating the Dart Model from One Fact

IKey Formulas

FormulaMeaningWhat you need
P(r1<r<r2)=∫r1r22πc r e−r2 dr∫0∞2πc r e−r2 dr=e−r12−e−r22\displaystyle P(r_1 < r < r_2) = \frac{\int_{r_1}^{r_2} 2\pi c\,r\,e^{-r^2}\,dr}{\int_0^{\infty} 2\pi c\,r\,e^{-r^2}\,dr} = e^{-r_1^2} - e^{-r_2^2}Probability of a ring: part over whole, with weight w(r)=2πc r e−r2w(r) = 2\pi c\,r\,e^{-r^2}The two radii; the whole is cπc\pi, so cc cancels
1−e−a2=12  ⇒  e−a2=121 - e^{-a^2} = \tfrac12 \;\Rightarrow\; e^{-a^2} = \tfrac12Calibration: the thrower hits the target of radius aa half the timeOne measured hit rate; aa itself is never needed
e−(ka)2=(e−a2)k2=(12)k2e^{-(ka)^2} = \left(e^{-a^2}\right)^{k^2} = \left(\tfrac12\right)^{k^2}Every circle at a multiple of aa becomes a power of one halfThe radius as a multiple kaka of the target radius
P(brother)=16[(12)4−(12)9]=313072≈1%P(\text{brother}) = \tfrac16\left[\left(\tfrac12\right)^4 - \left(\tfrac12\right)^9\right] = \tfrac{31}{3072} \approx 1\%The brother's piece: ring 2a2a to 3a3a, from 3 to 5 o'clockTwo hours of twelve =16= \tfrac16 of the ring

Key Insight: The model has one unknown scale, and one fact about the thrower fixes it: e−a2=12e^{-a^2} = \tfrac12. After that, any ring whose radii are multiples of aa has probability equal to a difference of powers of one half.

IIVisualization 1 — The Brother's Piece of the Board

With e−a2=12e^{-a^2} = \tfrac12, the ring from kaka to mama has probability (12)k2−(12)m2\left(\tfrac12\right)^{k^2} - \left(\tfrac12\right)^{m^2}, and a clock sector takes its share.

IIIVisualization 2 — Three Thousand Throws

Darts drawn from the density c e−r2c\,e^{-r^2} hit the target about half the time and the brother about once per hundred throws.

IVVisualization 3 — The Weight w(r)w(r)

The density e−r2e^{-r^2} is highest at the bullseye, but the weight w(r)=2πc r e−r2w(r) = 2\pi c\,r\,e^{-r^2}, hits per unit of rr, is zero there.

💡 Challenge: find the band of width 0.250.25 that is most likely to be hit; it is not the one at the bullseye.

VQuiz Questions

Problem 1 · The First Ring Outside the Target

Given: Ralph's thrower, with e−a2=12e^{-a^2} = \tfrac12 — find the probability that a dart lands between radius aa and radius 2a2a.

✅ Correct! e−a2−e−4a2=12−(12)4=12−116=716e^{-a^2} - e^{-4a^2} = \tfrac12 - \left(\tfrac12\right)^4 = \tfrac12 - \tfrac1{16} = \tfrac{7}{16}.
❌ The power is the square of the multiple. e−(2a)2=e−4a2=(12)4e^{-(2a)^2} = e^{-4a^2} = \left(\tfrac12\right)^4, not (12)2\left(\tfrac12\right)^2.
❌ That includes the target. 1−116=15161 - \tfrac1{16} = \tfrac{15}{16} is P(r<2a)P(r < 2a); the target disc, worth 12\tfrac12, lies inside radius aa and is not part of the ring.
❌ That is the region beyond 2a2a. e−4a2=116e^{-4a^2} = \tfrac1{16} is P(r>2a)P(r > 2a); the ring needs e−a2e^{-a^2} minus it.
❌ Not quite. Use P(r1<r<r2)=e−r12−e−r22P(r_1 < r < r_2) = e^{-r_1^2} - e^{-r_2^2} with r1=ar_1 = a, r2=2ar_2 = 2a.
Show solution
P(a<r<2a)=e−a2−e−(2a)2=e−a2−(e−a2)4=12−116=716.P(a < r < 2a) = e^{-a^2} - e^{-(2a)^2} = e^{-a^2} - \left(e^{-a^2}\right)^4 = \frac12 - \frac1{16} = \frac{7}{16}.

Together with the target's 12\tfrac12 and the 116\tfrac1{16} beyond 2a2a, the three pieces add up to 816+716+116=1\tfrac{8}{16} + \tfrac{7}{16} + \tfrac{1}{16} = 1.

Problem 2 · Calibrating a Better Thrower

Given: a different thrower, with the same model c e−r2c\,e^{-r^2}, hits the target of radius aa three times out of four — find e−a2e^{-a^2} for this thrower.

✅ Correct! 1−e−a2=341 - e^{-a^2} = \tfrac34 gives e−a2=14e^{-a^2} = \tfrac14: the chance of missing the target.
❌ That sets e−a2e^{-a^2} equal to the hit rate. The hit rate is P(0<r<a)=1−e−a2P(0 < r < a) = 1 - e^{-a^2}; e−a2e^{-a^2} is the miss rate.
❌ ln⁡4\ln 4 is a2a^2, not e−a2e^{-a^2}. Solving e−a2=14e^{-a^2} = \tfrac14 for a2a^2 is a step the calculation never needs; the value of e−a2e^{-a^2} itself is what goes into every ring.
❌ ln⁡4\sqrt{\ln 4} is the radius aa itself. No probability needs aa; it needs e−a2e^{-a^2}, and that number is less than 11.
❌ Not quite. Write the hit rate as P(0<r<a)P(0 < r < a) and set it equal to 34\tfrac34.
Show solution
P(0<r<a)=e−02−e−a2=1−e−a2=34⟹e−a2=14.P(0 < r < a) = e^{-0^2} - e^{-a^2} = 1 - e^{-a^2} = \frac34 \quad\Longrightarrow\quad e^{-a^2} = \frac14.

A more accurate thrower has a larger target relative to the spread of the darts, so e−a2e^{-a^2} is smaller: 14\tfrac14 instead of Ralph's 12\tfrac12.

Problem 3 · The Brother and the Better Thrower

Given: the thrower of Problem 2, with e−a2=14e^{-a^2} = \tfrac14, and the brother in the same place, the ring from 2a2a to 3a3a between 3 and 5 o'clock — find his chance of being hit on one throw.

✅ Correct! 16[(14)4−(14)9]≈16⋅1256≈0.065%\tfrac16\left[\left(\tfrac14\right)^4 - \left(\tfrac14\right)^9\right] \approx \tfrac16 \cdot \tfrac1{256} \approx 0.065\%, about one throw in 15001500.
❌ That is the answer for Ralph's thrower. With e−a2=14e^{-a^2} = \tfrac14 the powers are of 14\tfrac14, not 12\tfrac12.
❌ That is the whole ring. (14)4−(14)9≈0.0039\left(\tfrac14\right)^4 - \left(\tfrac14\right)^9 \approx 0.0039; the brother covers only 22 of the 1212 hours, so take 16\tfrac16 of it.
❌ The powers are k2k^2, not kk. 16[(14)2−(14)3]=1128\tfrac16\left[\left(\tfrac14\right)^2 - \left(\tfrac14\right)^3\right] = \tfrac1{128} uses e−2a2e^{-2a^2} and e−3a2e^{-3a^2}; the radii 2a2a and 3a3a give e−4a2e^{-4a^2} and e−9a2e^{-9a^2}.
❌ Not quite. Ring probability (e−a2)4−(e−a2)9\left(e^{-a^2}\right)^4 - \left(e^{-a^2}\right)^9, then 16\tfrac16 of it.
Show solution

The ring from 2a2a to 3a3a:

e−4a2−e−9a2=(14)4−(14)9=1256−1262144=1023262144.e^{-4a^2} - e^{-9a^2} = \left(\tfrac14\right)^4 - \left(\tfrac14\right)^9 = \frac{1}{256} - \frac{1}{262144} = \frac{1023}{262144}.

The brother's two hours of twelve take one sixth of it:

P(brother)=16⋅1023262144=341524288≈0.00065=0.065%.P(\text{brother}) = \frac16 \cdot \frac{1023}{262144} = \frac{341}{524288} \approx 0.00065 = 0.065\%.

The better thrower makes the brother about 1515 times safer than with Ralph's 313072≈1.01%\tfrac{31}{3072} \approx 1.01\%.

Problem 4 · The Most Likely Distance

Given: the weight w(r)=2πc r e−r2w(r) = 2\pi c\,r\,e^{-r^2} for r≥0r \ge 0 — find the radius where ww is largest.

✅ Correct! w′(r)=2πc (1−2r2) e−r2=0w'(r) = 2\pi c\,(1 - 2r^2)\,e^{-r^2} = 0 at r2=12r^2 = \tfrac12. Since a=ln⁡2≈0.83a = \sqrt{\ln 2} \approx 0.83, the peak r≈0.71≈0.85ar \approx 0.71 \approx 0.85a sits just inside the edge of the target.
❌ That is where e−r2e^{-r^2} peaks, the most likely single spot. The weight has the extra factor rr from the shell circumference 2πr2\pi r, so w(0)=0w(0) = 0.
❌ Check the chain rule. ddre−r2=−2r e−r2\tfrac{d}{dr}e^{-r^2} = -2r\,e^{-r^2}; dropping the 2r2r gives (1−r) e−r2(1 - r)\,e^{-r^2} and the wrong root r=1r = 1.
❌ The critical point solves 1−2r2=01 - 2r^2 = 0. That gives r2=12r^2 = \tfrac12, so r=12r = \tfrac{1}{\sqrt2}, not 12\tfrac12.
❌ Not quite. Differentiate r e−r2r\,e^{-r^2} with the product rule and set the result to zero.
Show solution

Product rule and chain rule:

w′(r)=2πc(e−r2+r⋅(−2r) e−r2)=2πc (1−2r2) e−r2.w'(r) = 2\pi c\left(e^{-r^2} + r \cdot (-2r)\,e^{-r^2}\right) = 2\pi c\,(1 - 2r^2)\,e^{-r^2}.

Since e−r2>0e^{-r^2} > 0, w′(r)=0w'(r) = 0 only when 1−2r2=01 - 2r^2 = 0, so r=12r = \tfrac{1}{\sqrt2}. For smaller rr, w′>0w' > 0; for larger rr, w′<0w' < 0: a maximum.

Spot by spot the bullseye is most likely, but the most likely distance from the centre is 12\tfrac{1}{\sqrt2}: near the centre there is too little area, far out the density is too small.

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