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To return to the original price after 2 days, the stock must move up 1 unit and then down 1 unit, or vice versa. The two scenarios are independent, so we can solve the problem by calculating the probability of each scenario and then adding them up. Scenario 1: The stock moves up 1 unit in the first day and down 1 unit in the second day. Probability of this scenario = \(p \cdot (1-p)\) Scenario 2: The stock moves down 1 unit in the first day and up 1 unit in the second day. Probability of this scenario = \((1-p) \cdot p\) The total probability of the stock being at its original price after 2 days is the sum of the probabilities of both scenarios: Probability of stock at original price after 2 days = \(p \cdot (1-p) + (1-p) \cdot p = 2p(1-p)\).

There are 3 days and 2 possible outcomes (up or down) for each day, so there are 2^3 = 8 possible ways the stock's price changes: 1- UUU: up, up, up 2- UUD: up, up, down 3- UDU: up, down, up 4- DDU: down, down, up 5- DUU: down, up, up 6- DUD: down, up, down 7- UDD: up, down, down 8- DDD: down, down, down For the stock's price to increase by 1 unit after 3 days, there must be two ups and one down: UUD (Scenario 1), UDU (Scenario 2), and DUU (Scenario 3). The probability of each scenario: - Scenario 1: \(p \cdot p \cdot (1-p) = p^2(1-p)\) - Scenario 2: \(p \cdot (1-p) \cdot p = p^2(1-p)\) - Scenario 3: \((1-p) \cdot p \cdot p = p^2(1-p)\) The total probability of the stock's price increased by 1 unit after 3 days is the sum of the probabilities of the three scenarios: Probability of stock price increased by 1 unit after 3 days = \(3p^2(1-p)\).

We can use conditional probability to solve this problem. Let A be the event that the stock went up on the first day. Let B be the event that the stock's price increased by 1 unit after 3 days. We are asked to find the probability of A, given B, which is written as P(A|B). By definition, the conditional probability P(A|B) can be written as: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) We already calculated the probability of event B in Part (b), which is \(P(B) = 3p^2(1-p)\). Now, we need to find the probability of both events A and B occurring, which can be represented by the intersection of A and B, or P(A ∩ B). From the scenarios discussed in Part (b), only scenario 1 and 2 involve the stock going up on the first day (event A) and the stock's final price increasing by 1 unit after 3 days (event B). - Scenario 1: \(p \cdot p \cdot (1-p) = p^2(1-p)\) - Scenario 2: \(p \cdot (1-p) \cdot p = p^2(1-p)\) The probability of (A ∩ B) is the sum of the probabilities of scenario 1 and 2: \(P(A \cap B) = 2p^2(1-p)\) Now, we can plug these probabilities into the conditional probability formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2p^2(1-p)}{3p^2(1-p)} = \frac{2}{3}\) The probability that the stock went up on the first day, given that after 3 days the stock's price increased by 1 unit, is \(\frac{2}{3}\).