Bayes’ Theorem and Examples

Formula

The formula for Bayes’ Theorem is given by:

$$P(E_i|A)=\frac{P(E_i)P(A|E_i)}{\sum _{j=1}^{n}P(E_j)P(A|E_j)}$$

Key Terminology

• $E_i$ are hypotheses or possible causes.
• $P(E_i)$ is the prior probability of $E_i$.
• $P(E_i|A)$ is the posterior probability of $E_i$.
• The denominator ensures normalization over all possible hypotheses.

Example 1: Probability of a Red Ball from Bag II

Problem: Suppose we have two bags:

• Bag I: 3 red, 4 black
• Bag II: 5 red, 6 black
• A bag is randomly chosen, and a red ball is drawn. Find the probability that it was from Bag II.

Step 1: Define Probabilities

$$P(B_1)=\frac{1}{2},P(B_2)=\frac{1}{2}$$

$$P(R|B_1)=\frac{3}{7},P(R|B_2)=\frac{5}{11}$$

Step 2: Apply Bayes’ Theorem

$$P(B_2|R)=\frac{P(B_2)P(R|B_2)}{P(B_1)P(R|B_1)+P(B_2)P(R|B_2)}$$

$$=\frac{\frac{1}{2}\times \frac{5}{11}}{\frac{1}{2}\times \frac{3}{7}+\frac{1}{2}\times \frac{5}{11}}=\frac{5}{8}$$

Example 2: Probability of Another Gold Coin

Problem: Three boxes contain:

• Box I: 2 gold coins
• Box II: 2 silver coins
• Box III: 1 gold, 1 silver
• A box is randomly selected, and a gold coin is drawn. Find the probability that the other coin is also gold.

Step 1: Assign Probabilities

$$P(B_1)=P(B_2)=P(B_3)=\frac{1}{3}$$

$$P(G|B_1)=1,P(G|B_2)=0,P(G|B_3)=\frac{1}{2}$$

Step 2: Apply Bayes’ Theorem

$$P(B_1|G)=\frac{P(B_1)P(G|B_1)}{P(B_1)P(G|B_1)+P(B_3)P(G|B_3)}$$

$$=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{1}{2}}=\frac{2}{3}$$

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