Bayes'Rules

• Our sample space $S$ is the population of adults in a small town. They can be categorized according to employment status.
• One individual is to be selected at random for a publicity tour.
  • The concerned event $E$: the one chosen is employed
  • Give the additional information that 36 of those employed and 12 of those unemployed are members of the Rotary Club.
  • Find the probability of the event $A$ that individual selected is a member of the Rotary Club.

Figure 4: Venn diagram for the events $A$, $E$, and $E'$.

• Event $A$ is the union of the two mutually exclusive events $E \cap A$ and $E' \cap A$. Hence,
  $$\bullet A=(E \cap A)\cup (E' \cap A)$$
  $$\bullet P(A)=P[(E \cap A)\cup (E' \cap A)]$$
  $$=P(E \cap A) + P(E' \cap A)$$
  $$=P(E)P(A\vert E) + P(E')P(A\vert E')$$
  $$\bullet P(E)=\frac{600}{900}=\frac{2}{3},~P(A\vert E)=\frac{36}{600}=\frac{3}{50}$$
  $$\bullet P(E')=\frac{1}{3},P(A\vert E)=\frac{12}{300}=\frac{1}{25}$$
  $$\bullet P(A)=\frac{2}{3}*\frac{3}{50}+\frac{1}{3}*\frac{1}{25}=\frac{4}{75}$$

Figure 5: Tree diagram for the data.

• Theorem 2.16: (Theorem of total probability or rule of elimination)
  

Figure 6: Partitioning the sample space S.

• Example 2.41: In a certain assembly plant, three machines, $B_1$, $B_2$ and $B_3$ make 30%, 45% and 25%, respectively, of the products.
• It is known from past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective.
• Now, suppose that a finished product is randomly selected. What is the probability that it is defective?
  $$P(A)=P(B_1)P(A\vert B_1)+P(B_2)P(A\vert B_2)+P(B_3)P(A\vert B_3)$$
  $$=03*0.02+0.45*0.03+0.25*0.02=0.0245$$
  
  

  $\bullet$ Solution:
  $\bullet$ Event $A$: the product is defective.
  $\bullet$ Event $B$: the product is made by machine $B_i$

  Figure 7: Tree diagram for Example 2.41.

  

• Theorem 2.17: (Bayes'Rule)
  
• It can be proved by the definition of conditional probability,
  $$P(B_r\vert A) = P(B_r \cap A) /P(A)$$
  and then using Theorem 2.16 in the denominator.
• Useful in problems where $P(B_i\vert A)$ are not known but $P(A\vert B_i)$ and $P(B_i)$ are known.
• Some terminology:
  • $P(B_i)$ : priors
  • $P(A\vert B_i)$ : likelihoods
  • $P(B_i\vert A)$ : posteriors

• Example 2.42: With reference to Example 2.41, if a product were chosen randomly and found to be defective, what is the probability that it was made by machine $B_3$
• Using Bayes'rule,
  $$P(B_3\vert A) =\frac{P(B_3)P(A\vert B_3)}{P(B_1)P(A\vert B_1)+ P(B_2)P(A\vert B_2)+P(B_3)P(A\vert B_3)}$$
  $$=\frac{0.005}{0.006+0.0135+0.005}=\frac{10}{49}$$

• Example 2.43: A manufacturing firm employs three analytical plans for the design and development of a particular product.
• For cost reasons, all three are used at varying times. In fact, plans 1, 2, and 3 are used for 30%, 20%, and 50% of the products respectively.
• The ``defect rate'' is different for the three procedures as follows:
  $$P(D\vert P_1)=0.01,~P(D\vert P_2)=0.03,~P(D\vert P_3)=0.5$$
  where $P(D\vert P_j)$ is the probability of a defective product, given plan $j$.
• If a random product was observed and found to be defective, which plan was most likely used and thus responsible?
• Solution: $P(P_1)=0.3,(P_12)=0.2,(P_1)=0.5$
  $$P(P_i\vert D)=\frac{P(P_i)P(D\vert P_i)}{\sum_{i=1}^3 P(P_i)P(D\v... ...c{(0.30)(0.01)}{(0.3)(0.01) + (0.20)(0.03) + (0.50)(0.02)}=\frac{0.003}{0.019}$$
  $$P(P_1\vert D)=0.158,~P(P_2\vert D)=0.316,~ P(P_3\vert D)=0.526.$$