Porisma

Worked solution

Distinguish a reflected distribution from dependent realised throws

A cubical die has three faces marked 1, two faces marked 2 and one face marked 3. Find the expectation and variance of its score in one throw.

Deduce, or find otherwise, the expectation and variance for a second cubical die with one face marked 1, two marked 2 and three marked 3.

Two dice of the first type and one of the second are thrown together; X is their total score. Let μ=E(X)\mu=E(X) and σ2=Var⁡(X)\sigma^2=\operatorname{Var}(X). Show that

(i) σ2=5/3\sigma^2=5/3,

(ii) P(∣X−μ∣>2σ)=1/18P(|X-\mu|>2\sigma)=1/18.

[J1983/I/11]

CJC · Discussion · Question 6

Answer

E(U)=5/3E(U)=5/3, Var⁡(U)=5/9\operatorname{Var}(U)=5/9; E(V)=7/3E(V)=7/3, Var⁡(V)=5/9\operatorname{Var}(V)=5/9; μ=17/3\mu=17/3, σ2=5/3\sigma^2=5/3; P(∣X−μ∣>2σ)=1/18P(|X-\mu|>2\sigma)=1/18.

First-die distribution and source model

Let U be a score on the first type. The source treats the six physical faces as equally likely and the throws as independent.

uP(U=u)P(U=u)
13/63/6
22/62/6
31/61/6
How did this part go?

First-die moments

E(U)=1(3/6)+2(2/6)+3(1/6)=5/3E(U)=1(3/6)+2(2/6)+3(1/6)=5/3. Also E(U2)=12(3/6)+22(2/6)+32(1/6)=10/3E(U^2)=1^2(3/6)+2^2(2/6)+3^2(1/6)=10/3. Hence Var⁡(U)=10/3−(5/3)2=5/9\operatorname{Var}(U)=10/3-(5/3)^2=5/9.

How did this part go?

Second die, source method 1: deduce by reflecting the face labels

The source lists u:1,1,1,2,2,3 and corresponding v:3,3,3,2,2,1, writing V=4−U. Precisely, V has the same distribution as 4−U. Thus E(V)=E(4)−E(U)=4−5/3=7/3E(V)=E(4)-E(U)=4-5/3=7/3, and Var⁡(V)=Var⁡(4−U)=(−1)2Var⁡(U)=Var⁡(4)+Var⁡(U)=0+5/9=5/9\operatorname{Var}(V)=\operatorname{Var}(4-U)=(-1)^2\operatorname{Var}(U)=\operatorname{Var}(4)+\operatorname{Var}(U)=0+5/9=5/9. This moment deduction is not a pointwise relation between independent dice thrown later.

How did this part go?

Second die, source method 2: direct distribution and moments

vP(V=v)P(V=v)
11/61/6
22/62/6
33/63/6

E(V)=1(1/6)+2(2/6)+3(3/6)=7/3E(V)=1(1/6)+2(2/6)+3(3/6)=7/3. Also E(V2)=12(1/6)+22(2/6)+32(3/6)=6E(V^2)=1^2(1/6)+2^2(2/6)+3^2(3/6)=6. Therefore Var⁡(V)=6−(7/3)2=5/9\operatorname{Var}(V)=6-(7/3)^2=5/9.

How did this part go?

(i) Add means and independent variances

For independent throws X=U1+U2+V1X=U_1+U_2+V_1. Then μ=E(U1)+E(U2)+E(V1)=5/3+5/3+7/3=17/3\mu=E(U_1)+E(U_2)+E(V_1)=5/3+5/3+7/3=17/3. Independence gives σ2=Var⁡(U1)+Var⁡(U2)+Var⁡(V1)=5/9+5/9+5/9=15/9=5/3\sigma^2=\operatorname{Var}(U_1)+\operatorname{Var}(U_2)+\operatorname{Var}(V_1)=5/9+5/9+5/9=15/9=5/3.

How did this part go?

(ii) Split the absolute-deviation event into two tails

P(∣X−μ∣>2σ)=P ⁣(X−173>253)+P ⁣(X−173<−253).\begin{aligned}&P(|X-\mu|>2\sigma)\\ & =P\!\left(X-\frac{17}3>2\sqrt{\frac53}\right)\\ &{}+P\!\left(X-\frac{17}3<-2\sqrt{\frac53}\right).\end{aligned}

The source rounds the corresponding cutoffs to 8.25 and 3.08. Since X is an integer from 3 to 9, the exact thresholds imply the same events X=9X=9 or X=3X=3. This retains the two tails; extraction loses the printed absolute-value bars.

How did this part go?

(ii) Retain the two extreme-outcome products

P(X=9)+P(X=3)=P(U1=3)P(U2=3)P(V1=3)+P(U1=1)P(U2=1)P(V1=1)=(1/6)(1/6)(1/2)+(1/2)(1/2)(1/6)=1/18P(X=9)+P(X=3)=P(U_1=3)P(U_2=3)P(V_1=3)+P(U_1=1)P(U_2=1)P(V_1=1)=(1/6)(1/6)(1/2)+(1/2)(1/2)(1/6)=1/18.

How did this part go?

Common mistakes

The source’s V=4−U is a distributional deduction, not a dependence relation between actual independent throws. The tail event uses absolute value and σ, not σ².

Solution notes

Both initial unnumbered expectation/variance asks and both numbered show-that asks retained. Both second-die methods, full face lists/tables, every moment calculation, sum/variance stage, absolute-value bars, exact square-root thresholds, source rounded cutoffs and two extreme products retained. V=4−U qualified as equality in distribution; equal physical-face probabilities and independent throws made explicit. No marks printed.

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