STA256 Lecture 04

• Law of Total Probability
• Conditional Probability
  • Theorem 20
  • We have $\{B_1,B_2,\dots ,B_k\}$
  • $A=A\cap B_1\cup A\cap B_2\cup \cdots \cup A\cap B_k$
  • $P(A)=P(A\cap B_1\cup A\cap B_2\cup \cdots \cup A\cap B_k)$
  • Since the $B_i$ are disjoint, we can just add the probability
  • $P(A)=P(A\cap B_1)+P(A\cap B_2)+\cdots +P(A\cap B_k)$
  • We can use the Conditional Probability rules to sub in a formula
  • $P(A)=P(B_1)\cdot P(A|B_1)+P(B_2)\cdot P(A|B_2)+\cdots +P(B_k)\cdot P(A|B_k)$
  • So this is $\sum _{i=1}^k{B}_i\cdot P(A|B_i)$
• Bayes' Theorem