Porisma

Worked solution

Count tries when keys are tested without replacement

John has four keys, only one of which unlocks a door. He tries the keys randomly without replacement. X is the number of tries needed to unlock the door.

(i) Show that P(X=3)=1/4P(X=3)=1/4.

(ii) Tabulate the distribution of X.

(iii) Write down E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

(iv) Find P(X2<8)P(X^2<8).

[2006/CJC/II/23]

CJC · Lecture · Example 16

(i) Follow the unsuccessful, unsuccessful, successful path

After each failed key is removed, the number of keys changes. Thus P(X=3)=(3/4)(2/3)(1/2)=1/4P(X=3)=(3/4)(2/3)(1/2)=1/4. This makes the probability/counting scenario explicit, as the Teacher note requests.

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(ii) Tabulate the number of tries

The successful key is equally likely to occupy each position in the random testing order (an explanation of the source uniform law). In particular, the source calculates P(X=2)=(3/4)(1/3)=1/4P(X=2)=(3/4)(1/3)=1/4.

xx1234
P(X=x)P(X=x)1/41/41/41/41/41/41/41/4
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(iii) Calculate the mean and variance

By symmetry E(X)=(2+3)/2=5/2E(X)=(2+3)/2=5/2. The source second moment is E(X2)=12(1/4)+22(1/4)+32(1/4)+42(1/4)=15/2E(X^2)=1^2(1/4)+2^2(1/4)+3^2(1/4)+4^2(1/4)=15/2. Therefore Var⁡(X)=E(X2)−[E(X)]2=15/2−(2.5)2=5/4\operatorname{Var}(X)=E(X^2)-[E(X)]^2=15/2-(2.5)^2=5/4.

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(iv) Split the squared inequality

Follow the Teacher emphasis: X2<8X^2<8 means −8<X<8-\sqrt8<X<\sqrt8. Of the possible values 1,2,3,41,2,3,4, only 1 and 2 qualify. Hence P(X2<8)=P(X=1)+P(X=2)=1/4+1/4=1/2P(X^2<8)=P(X=1)+P(X=2)=1/4+1/4=1/2. This completes the Student blank using the Teacher working.

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Common mistakes

The key is not replaced after failure. Do not use a geometric law with a constant success probability. Split a squared inequality into both bounds before checking the support.

Read the hint