Extra Probability (HL)
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
The discrete random variable \(X\) has \(E(X)=4\) and \(\text{Var}(X)=5\). Find:
(a)\(E(3X-2)\)
(b)\(\text{Var}(3X-2)\)
(c)\(E(2-X)\)
(d)\(\text{Var}(2-X)\)
Linear transformation: \(E(aX+b)=aE(X)+b\), \(\text{Var}(aX+b)=a^2\text{Var}(X)\)
(a)
Answer
\(E(3X-2)=3(4)-2=10\)
(b)
Answer
\(\text{Var}(3X-2)=3^2(5)=45\)
(c)
Answer
\(E(2-X)=-1(4)+2=-2\)
(d)
Answer
\(\text{Var}(2-X)=(-1)^2(5)=5\)
Question 2
Events A and B satisfy \(P(A)=\tfrac{1}{3}\), \(P(B|A)=\tfrac{1}{2}\), \(P(B|A')=\tfrac{1}{4}\). Find \(P(A|B)\).
Total probability
\(P(B)=\tfrac{1}{3}\times\tfrac{1}{2}+\tfrac{2}{3}\times\tfrac{1}{4}=\tfrac{1}{6}+\tfrac{1}{6}=\tfrac{1}{3}\)
Bayes
\(P(A|B)=\dfrac{P(B|A)P(A)}{P(B)}=\dfrac{\tfrac{1}{2}\times\tfrac{1}{3}}{\tfrac{1}{3}}\)
Answer
\(P(A|B)=\dfrac{1}{2}\)
Question 3
A continuous random variable \(X\) has PDF \(f(x)=\lambda e^{-\lambda x}\) for \(x\geq0\) (Exponential distribution). Show that \(E(X)=\dfrac{1}{\lambda}\) and find the median.
\(E(X)=\displaystyle\int_0^\infty x\lambda e^{-\lambda x}\,dx\)
IBP with \(u=x,\;dv=\lambda e^{-\lambda x}\): \([-xe^{-\lambda x}]_0^\infty+\displaystyle\int_0^\infty e^{-\lambda x}\,dx=0+[-\tfrac{1}{\lambda}e^{-\lambda x}]_0^\infty=\tfrac{1}{\lambda}\) ✓
Median: solve \(\displaystyle\int_0^m\lambda e^{-\lambda x}\,dx=\tfrac{1}{2}\)
\(1-e^{-\lambda m}=\tfrac{1}{2}\Rightarrow m=\dfrac{\ln2}{\lambda}\)
Answer
Median \(=\dfrac{\ln2}{\lambda}\)
Question 4
A continuous random variable \(X\) has PDF \(f(x)=3x^2\) for \(0\leq x\leq1\) (and \(f(x)=0\) otherwise). Find \(E(X)\) and \(\text{Var}(X)\).
\(E(X)=\displaystyle\int_0^1 x\cdot3x^2\,dx=\int_0^13x^3\,dx=\left[\tfrac{3x^4}{4}\right]_0^1\)
\(E(X^2)=\displaystyle\int_0^1 x^2\cdot3x^2\,dx=\int_0^13x^4\,dx=\left[\tfrac{3x^5}{5}\right]_0^1\)
\(\text{Var}(X)=E(X^2)-[E(X)]^2=\tfrac{3}{5}-\left(\tfrac{3}{4}\right)^2=\tfrac{3}{5}-\tfrac{9}{16}\)
Answer
\(E(X)=\tfrac{3}{4}\); \(\text{Var}(X)=\tfrac{3}{80}\;(=0.0375)\)
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