Statistics & Probability — Intermediate

Question 1

(a) Derivation. (4)

By definition of conditional probability (from Kolmogorov axioms, with $P(D)>0$): $P(S\mid D)=\frac{P(S\cap D)}{P(D)},\qquad P(S\cap D)=P(D\mid S)P(S).\ \text{(1 mark)}$ By the law of total probability, since $\{S,B\}$ partition the sample space ($B=S^c$): $P(D)=P(D\mid S)P(S)+P(D\mid B)P(B).\ \text{(2 marks)}$ Substituting: $P(S\mid D)=\frac{P(D\mid S)P(S)}{P(D\mid S)P(S)+P(D\mid B)P(B)}.\ \text{(1 mark)}$

(b) (4) With $P(B)=0.999$:

• Numerator $=0.98\times0.001 = 0.00098$. (1)
• Denominator $=0.00098 + 0.03\times0.999 = 0.00098 + 0.02997 = 0.03095$. (1)
• $P(S\mid D)=0.00098/0.03095 = 0.03167\approx \mathbf{0.0317}$ (3 s.f.). (1)

Interpretation: Signals are extremely rare ($0.1\%$), so even a small false-alarm rate ($3\%$) applied to the vast background produces far more false firings ($\approx0.03$) than true firings ($\approx0.001$). High efficiency cannot overcome a low base rate — purity is dominated by the prior. (1)

(c) (5) Conditional independence assumption: given the event type ($S$ or $B$), $D_1$ and $D_2$ are independent, so $P(D_1\cap D_2\mid S)=P(D\mid S)^2=0.98^2=0.9604,$ $P(D_1\cap D_2\mid B)=P(D\mid B)^2=0.03^2=0.0009.$ (2, for stating & using assumption)

Bayes again: $P(S\mid D_1D_2)=\frac{0.9604\times0.001}{0.9604\times0.001+0.0009\times0.999}=\frac{0.0009604}{0.0009604+0.0008991}.$ (2) Denominator $=0.0018595$; ratio $=0.0009604/0.0018595=\mathbf{0.5165}\approx0.516$. (1) (Coincidence requirement dramatically raises purity from 3% to ~52%.)

(d) (7) Logic: for $n$ identical conditionally-independent detectors all firing, $P(S\mid n)=\frac{(\text{eff})^n\cdot\text{prior}}{(\text{eff})^n\cdot\text{prior}+(\text{far})^n\cdot(1-\text{prior})}.$

def posterior(prior, eff, far, n_fires):
    like_s = eff**n_fires * prior
    like_b = far**n_fires * (1 - prior)
    return like_s / (like_s + like_b)
 
n = 1
while posterior(0.001, 0.98, 0.03, n) < 0.99:
    n += 1
print(n, posterior(0.001, 0.98, 0.03, n))

Checking:

• $n=2$: $0.516$;
• $n=3$: like_s $=0.98^3(0.001)=9.412\times10^{-4}$, like_b $=0.03^3(0.999)=2.697\times10^{-5}$; posterior $=9.412\text{e-}4/(9.412\text{e-}4+2.697\text{e-}5)=0.9721$.
• $n=4$: like_b $=0.03^4(0.999)=8.09\times10^{-7}$, like_s $=0.98^4(0.001)=9.224\text{e-}4$; posterior $=0.99912\ge0.99$. ✓

Smallest $\mathbf{n=4}$. (Marks: function 3, recurrence logic 2, correct $n=4$ with working 2.)

Question 2

(a) (5) $E[X]=\sum_{k=0}^n k\binom nk p^k q^{n-k},\ q=1-p.$ $k=0$ term vanishes. Use $k\binom nk=n\binom{n-1}{k-1}$ (1): $=\sum_{k=1}^n n\binom{n-1}{k-1}p^k q^{n-k}=np\sum_{k=1}^n\binom{n-1}{k-1}p^{k-1}q^{n-k}.\ (2)$ Let $j=k-1$: $=np\sum_{j=0}^{n-1}\binom{n-1}{j}p^{j}q^{(n-1)-j}=np(p+q)^{n-1}=np.$ (2)

(b) (6) Compute $E[X(X-1)]=\sum k(k-1)\binom nk p^k q^{n-k}$. Using $k(k-1)\binom nk=n(n-1)\binom{n-2}{k-2}$ (2): $=n(n-1)p^2\sum_{k=2}^n\binom{n-2}{k-2}p^{k-2}q^{n-k}=n(n-1)p^2(p+q)^{n-2}=n(n-1)p^2.\ (2)$ Then $E[X^2]=E[X(X-1)]+E[X]=n(n-1)p^2+np$. (1) $\operatorname{Var}(X)=E[X^2]-(E[X])^2=n(n-1)p^2+np-n^2p^2=np-np^2=np(1-p).\ (1)$

(c) (5) $n=20,p=0.15,q=0.85$:

• $E[X]=20(0.15)=\mathbf{3}$. (1)
• $\operatorname{Var}=20(0.15)(0.85)=\mathbf{2.55}$. (1)
• $P(X\le2)=\sum_{k=0}^2\binom{20}{k}0.15^k0.85^{20-k}$:
  • $k=0$: $0.85^{20}=0.038760$
  • $k=1$: $20(0.15)(0.85^{19})=20(0.15)(0.045599)=0.136796$
  • $k=2$: $190(0.0225)(0.85^{18})=190(0.0225)(0.053646)=0.229334$
  • Sum $=0.404890\approx\mathbf{0.4049}$ (4 s.f.). (3)

(d) (6) $E[\hat p]=E[X/n]=np/n=p$ ⇒ unbiased. (2) $\operatorname{Var}(\hat p)=\operatorname{Var}(X)/n^2=np(1-p)/n^2=p(1-p)/n$, so SE $=\sqrt{p(1-p)/n}$. (2) Plug-in $\hat p=6/20=0.3$: SE $=\sqrt{0.3\times0.7/20}=\sqrt{0.0105}=0.1025\approx\mathbf{0.102}$. (1) Report: count rate $\hat p=0.30\pm0.10$ decays per window (i.e. state estimate with its standard error / confidence interval, noting the large relative uncertainty from small $n$). (1)

Question 3

(a) (6) Midpoints: 10,30,50,70,90; $f$: 6,14,20,12,8; $\sum f=60$. $\sum fx = 60+420+1000+840+720=3040$. Mean $=3040/60=\mathbf{50.67}$ keV. (2)

Cumulative freq: 6, 20, 40, 52, 60. $N/2=30$ ⇒ median class $40\text{–}60$ ($L=40$, $CF=20$, $f=20$, $h=20$). (2) $\text{Median}=40+\frac{30-20}{20}\times20=40+10=\mathbf{50}\text{ keV}.\ (2)$

(b) (5) $\sum fx^2 = 6(100)+14(900)+20(2500)+12(4900)+8(8100)$ $=600+12600+50000+58800+64800=186800$. Variance $=186800/60 - (50.667)^2 = 3113.33 - 2567.11 = 546.22$. (3) SD $=\sqrt{546.22}=23.37\approx\mathbf{23.4}$ keV (3 s.f.). (2)

(c) (4) $Q_1$: position $N/4=15$ → class $20\text{–}40$ ($L=20,CF=6,f=14,h=20$): $Q_1=20+\frac{15-6}{14}\times20=20+12.857=\mathbf{32.86}$ keV. (1.5) $Q_3$: position $3N/4=45$ → class $60\text{–}80$ ($L=60,CF=40,f=12,h=20$): $Q_3=60+\frac{45-40}{12}\times20=60+8.333=\mathbf{68.33}$ keV. (1.5) $\text{IQR}=68.33-32.86=\mathbf{35.48}$ keV. Box from 32.86 to 68.33, median line at 50 (≈ centred, slightly left), whiskers to 0 and 100. Since median ($50$) is near the box centre and mean ($50.67$)≈median, distribution is nearly symmetric with very slight right skew. (1)

(d) (3) Combinatorial proof: $\binom nr$ counts $r$-subsets of an $n$-set. Fix element $x$. Every $r$-subset either contains $x$ — choose the remaining $r-1$ from the other $n-1$ elements: $\binom{n-1}{r-1}$ ways — or excludes $x$ — choose all $r$ from the other $n-1$: $\binom{n-1}{r}$ ways. These cases are disjoint and exhaustive, so by the addition rule $\binom nr=\binom{n-1}{r-1}+\binom{n-1}{r}$. (2) This is exactly the rule that each entry in a Pascal-triangle row equals the sum of the two entries above it in the previous row, generating row $n$ from row $n-1$. (1)

[
  {"claim":"Q1b single-detector posterior ~0.0317",
   "code":"num=Rational(98,100)*Rational(1,1000); den=num+Rational(3,100)*Rational(999,1000); val=float(num/den); result = abs(val-0.0317)<0.001"},
  {"claim":"Q1d smallest n with posterior>=0.99 is 4",
   "code":"def post(n):\n    ls=(0.98**n)*0.001; lb=(0.03**n)*0.999; return ls/(ls+lb)\nn=1\nwhile post(n)<0.99:\n    n+=1\nresult = (n==4)"},
  {"claim":"Q2c P(X<=2)=0.4049 for Bin(20,0.15)",
   "code":"p=Rational(15,100); tot=sum(binomial(20,k)*p**k*(1-p)**(20-k) for k in range(3)); result = abs(float(tot)-0.4049)<0.0005"},
  {"claim":"Q3 mean=50.667, median=50, SD~23.4",
   "code":"import math\nf=[6,14,20,12,8]; x=[10,30,50,70,90]; N=sum(f)\nmean=sum(fi*xi for fi,xi in zip(f,x))/N\nvar=sum(fi*xi*xi for fi,xi in zip(f,x))/N-mean**2\nsd=math.sqrt(var)\nresult = abs(mean-50.667)<0.01 and abs(sd-23.37)<0.1"}
]