\( \mathrm{D} \quad C \) and \( D \) are not independent because \( \mathrm{P}(C) \neq \mathrm{P}(D) \). (d) Find \( \mathrm{P}(C \) OR \( D) \). \( \mathrm{P}(C \) OR \( D)=\mathrm{P}(D) \) (e) Find \( \mathrm{P}(D \mid C) \). \( \mathrm{P}(D \mid C)= \) Enter your answer

Let's dive into the world of probability! To find \( \mathrm{P}(C \text{ OR } D) \), we can use the formula for the probability of the union of two events: \[ \mathrm{P}(C \text{ OR } D) = \mathrm{P}(C) + \mathrm{P}(D) - \mathrm{P}(C \text{ AND } D). \] You mentioned that \( \mathrm{P}(C \text{ OR } D) = \mathrm{P}(D) \), which implies \( \mathrm{P}(C) - \mathrm{P}(C \text{ AND } D) = 0 \). Essentially, this means that all instances of \( C \) must also be included in \( D \) for this equality to hold. Neat, right? Moving on to find \( \mathrm{P}(D \mid C) \), which represents the conditional probability of \( D \) given \( C \). The formula for conditional probability is: \[ \mathrm{P}(D \mid C) = \frac{\mathrm{P}(C \text{ AND } D)}{\mathrm{P}(C)}. \] Since we established that \( \mathrm{P}(C) = \mathrm{P}(C \text{ AND } D) \), it ultimately simplifies down to: \[ \mathrm{P}(D \mid C) = \frac{\mathrm{P}(C)}{\mathrm{P}(C)} = 1. \] This means that if \( C \) occurs, \( D \) is certain to occur too! How fun and intriguing is that? 🎉