Study Group Worksheet Template

Probability · {course} · week {n}

One document, several people. Each problem is written by one person and read by a second, whose job is to say where the argument stopped being convincing rather than to fix it.

#	Topic	Written by	Checked by
1	Bayes, base rates	{name}	{name}
2	Expectation by conditioning	{name}	{name}
3	Normal approximation	{name}	{name}

1. A positive test result

A test is 99% sensitive and 95% specific. Prevalence is 1 in 1000. Someone tests positive.

Pr([+,D])=0.99,Pr([−,D^c])=0.95,Pr(D)=0.001

Pr([D,+])=(Pr([+,D])Pr(D))/(Pr([+,D])Pr(D)+Pr([+,D^c])Pr(D^c))

=((0.99)*(0.001))/((0.99)*(0.001)+(0.05)*(0.999))=0.00099/(0.00099+0.04995)=0.00099/0.05094≈0.0194

About 1.9%.

In 100,000 people there is roughly 1 true positive for every 50 false ones, because the 5% error rate acts on the 99.9% who are healthy while the 99% detection rate acts on the 0.1% who aren't. The test is good. The base rate is doing the damage.

Checker
The arithmetic holds. Worth adding: at prevalence 1 in 20 instead of 1 in 1000 the same test gives Pr([D,+])≈0.51. Same test, different population, opposite conclusion.

2. Rolls until the first six

X= number of rolls of a fair die until the first six, so X∼Geom*(1/6).

By conditioning on the first roll. Either it's a six, or you have started over having used one roll:

𝔼*[X]=1/6*(1)+5/6*(1+𝔼*[X])

𝔼*[X]=1+5/6*𝔼*[X]⟹1/6*𝔼*[X]=1⟹𝔼*[X]=6

Directly from the sum, with q=5/6 and p=1/6:

𝔼*[X]=(∑_k=1^∞)(k)*q^(k−1)*p=p*(∑_k=1^∞)(k)*q^(k−1)=p⋅1/((1−q)^2)=p/(p^2)=1/p=6

Both give 6. The conditioning argument is two lines and generalises to problems where the sum has no closed form, which is the reason to learn it.

Checker
The step 𝔼*[[X,first roll not a six]]=1+𝔼*[X] is memorylessness and should be named as such. It's the only real assumption in the argument.

3. 400 coin flips

X∼Bin*(400,1/2), so μ=200 and σ=√(,400⋅1/4)=10.

n*p=n*q=200, both far above 10, so the normal approximation is safe. With a continuity correction:

Pr(X>220)=Pr(X≥221)≈Pr(Z>(220.5−200)/10)=Pr(Z>2.05)≈0.0202

About 2%.

Without the continuity correction you get Pr(Z>2.00)≈0.0228, which is a 13% relative error on the tail. It matters most exactly here, in the tail, where the answer is small.

Checker
Agreed. Worth stating which theorem licenses this: de Moivre–Laplace, the CLT specialised to binomials.