STA256 Lecture 08

• Let $A,B$ be events in sample space $S$ where $P(A\cap B)=0$
• Disjoint
• Show that $P(A\cup B)=P(A)+P(B)$
• Notice the Addition Law of Probability
• States that it's $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
• we know $P(A\cap B)=0$
• so $P(A\cup B)=P(A)+P(B)-0⟹P(A)+P(B)$
• Solution:
  • By axiom 3
  • For pairwise disjoint events
  • $P\left(\bigcup _{i=1}^{\mathrm{\infty }}{A}_i\right)=\sum _{i=1}^{\mathrm{\infty }}P(A_i$
  • Here we have $P\left(\bigcup _{i=1}^2{A}_i\right)=\sum _{i=1}^{2}P(A_i)=P(A_1)+P(A_2)$
  • Label $A_1=A$ and $A_2=B$
  • So $P(A\cup B)=P(A)+P(B)$
• Review
  • Continuous Random Variable
    • Random Variables which can take any value in a range, larger support
  • Distribution Function
    • Probability Density Function
    • Cumulative Distribution Function
  • Percentile
• #tk slide 88
• Transformations - Univariate
• Section 8
  • Expected Value