Bayes' Theorem

• $$R_1$$: The first ball drawn is Red.
• $$B_1$$: The first ball drawn is Black.
• $$B_2$$: The second ball drawn is Black.

We need to find the conditional probability $$P(R_1 | B_2)$$. By Bayes' Theorem,

$$P(R_1 | B_2) = \frac{P(R_1) \cdot P(B_2 | R_1)}{P(B_2)}$$

Initially, the urn has 3 Red and 7 Black balls (Total = 10 balls).

• $$P(R_1) = \frac{3}{10}$$
• $$P(B_1) = \frac{7}{10}$$

• Case 1: First ball was Red ($$R_1$$)
  • One red ball is removed (leaving 2 Red, 7 Black).
  • Two additional red balls are added (leaving 2 + 2 = 4 Red, 7 Black).
  • Total balls now = 11.
  • $$P(B_2 | R_1) = \frac{7}{11}$$
• Case 2: First ball was Black ($$B_1$$)
  • One black ball is removed (leaving 3 Red, 6 Black).
  • Two additional black balls are added (leaving 3 Red, 6 + 2 = 8 Black).
  • Total balls now = 11.
  • $$P(B_2 | B_1) = \frac{8}{11}$$

Using the Law of Total Probability:

$$P(B_2) = P(R_1) \cdot P(B_2 | R_1) + P(B_1) \cdot P(B_2 | B_1)$$

$$P(B_2) = \left(\frac{3}{10} \times \frac{7}{11}\right) + \left(\frac{7}{10} \times \frac{8}{11}\right)$$

$$P(B_2) = \frac{21}{110} + \frac{56}{110} = \frac{77}{110}$$

$$P(R_1 | B_2) = \frac{\frac{21}{110}}{\frac{77}{110}}$$

$$P(R_1 | B_2) = \frac{21}{77} = \frac{3}{11}$$