Porisma

Worked solution

Use the absolute deviation and avoid double-counting the double-two case

X has the following distribution.

xP(X=x)P(X=x)
20.2
3a
4b
50.45

Given E(∣X−4∣)=11/10E(|X-4|)=11/10, find a and b. marks

Two independent observations of X are taken. Find the probability that one is 2 and the other is at most 4. marks

[2017/SRJC/II/5]

CJC · Review problems · Question 5

Answer

a=0.25a=0.25, b=0.1b=0.1; required probability 0.18.

Normalise

∑xP(X=x)=1\sum_xP(X=x)=1 gives 0.2+a+b+0.45=10.2+a+b+0.45=1, hence a+b=0.35a+b=0.35.

How did this part go?

Retain the full absolute-deviation equation

E(∣X−4∣)=∑x∣x−4∣P(X=x)=11/10E(|X-4|)=\sum_x|x-4|P(X=x)=11/10. Thus ∣2−4∣(0.2)+∣3−4∣a+∣4−4∣b+∣5−4∣(0.45)=11/10|2-4|(0.2)+|3-4|a+|4-4|b+|5-4|(0.45)=11/10. This reduces to 2(0.2)+a+0.45=11/102(0.2)+a+0.45=11/10, giving a=0.25 and b=0.1.

How did this part go?

Count both orders only for unequal observations

The event includes (2,2) once, and both orders of (2,3) and (2,4). Thus its probability is P(X1=2,X2=2)+2[P(X1=2,X2=3)+P(X1=2,X2=4)]=(0.2)(0.2)+2[(0.2)(0.25)+(0.2)(0.1)]=0.18P(X_1=2,X_2=2)+2[P(X_1=2,X_2=3)+P(X_1=2,X_2=4)]=(0.2)(0.2)+2[(0.2)(0.25)+(0.2)(0.1)]=0.18.

How did this part go?

Common mistakes

The event allows both observations to be 2; count this pair once, not twice.

Solution notes

Both asks/marks, full initial probability table, normalisation, all absolute-deviation terms and complete ordered-pair probability retained.

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