Paper 2 Section B Practice: Discrete Random Variables — Answers

Questions: Discrete Random Variables Drills.

Question 1

(a) Total probability gives

$$3a+b+0.25=1,$$

so $3a+b=0.75$. The mean gives

$$2a+2b+3(0.25)=1.65,$$

so $a+b=0.45$. Therefore

$$a=0.15,b=\mathrm{0.30.}$$

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(b)

$$P(0<X\le 2)=0.30+0.30=\mathrm{0.60.}$$

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(c)

$$E(X^2)=1^2(0.30)+2^2(0.30)+3^2(0.25)=\mathrm{3.75.}$$

Thus

$$\mathrm{Var}(X)=3.75-(1.65{)}^2=\mathrm{1.0275.}$$

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(d) For $1\le x<2$, the included values are $0$ and $1$, so

$$F(x)=0.15+0.30=\mathrm{0.45.}$$

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Question 2

(a)

$$E(S)=-2(0.3)+1(0.5)+4(0.2)=0.7,$$ $$E(S^2)=4(0.3)+1(0.5)+16(0.2)=\mathrm{4.9.}$$

Hence

$$\mathrm{Var}(S)=4.9-(0.7{)}^2=\mathrm{4.41.}$$

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(b)

$$E(R)=3E(S)+5=7.1,$$ $$\mathrm{Var}(R)=3^2\mathrm{Var}(S)=\mathrm{39.69.}$$

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(c)

$$E(T)=3E(S)=\mathrm{2.1.}$$

Since the plays are independent, their variances add:

$$\mathrm{Var}(T)=3\mathrm{Var}(S)=\mathrm{13.23.}$$

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(d) The sum is $3$ either when all three scores are $1$, or when the scores are $(-2,1,4)$ in any order. Therefore

$$P(T=3)=(0.5{)}^3+3!(0.3)(0.5)(0.2)=\mathrm{0.305.}$$

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