Cryptography probability

Question

62% of plaintext messages have even parity. 56% of odd plaintext messages have ciphertext with even parity. 48% of even plaintext messages have ciphertext with even parity. What is the probability that a ciphertext has even parity? I believe the answer is (.56)(.38) + (.62)(.48). Is this correct? What is the probability that even ciphertext decrypts to even plaintext? Is this question asking P(even plaintext | even ciphertext)? I think so but wanted some clarification. If yes, then would the answer be (.48)(.62) / first answer

Answer 1

62% of plaintext messages have even parity. 56% of odd plaintext messages have ciphertext with even parity. 48% of even plaintext messages have ciphertext with even parity.

Let $E$ be the event of the plaintext message having even parity.

Let $C$ be the event of the ciphertext having even parity.

Then we are given: $$\begin{align}\mathsf P(E) &= 0.62\\ \mathsf P(C\mid\neg E) &= 0.56\\ \mathsf P(C\mid E)&=0.48\end{align}$$

What is the probability that a ciphertext has even parity? I believe the answer is (.56)(.38) + (.62)(.48). Is this correct?

Yes. $$\begin{align} \mathsf P(C) & = \mathsf P(\neg E)\mathsf P(C\mid\neg E)+\mathsf P(E)\mathsf P(C\mid E) \\ & = (0.38)(0.56) + (0.62)(0.48) & \color{green}{\checkmark} \end{align}$$

What is the probability that even ciphertext decrypts to even plaintext? Is this question asking P(even plaintext | even ciphertext)? I think so but wanted some clarification. If yes, then would the answer be (.48)(.62) / first answer

$$\begin{align} \mathsf P(E\mid C) & = \frac{\mathsf P(E)\mathsf P(C\mid E)}{\mathsf P(C)} \\ & = \frac{(0.62)(0.48)}{(0.38)(0.56) + (0.62)(0.48)} & \color{green}{\checkmark} \end{align}$$