Question

Encryption systems with erroneous ciphertexts. In cryptography,ciphertext is encrypted or encoded text that is unreadable by a human or computer without the proper algorithm to decrypt it into plain text. The impact of erroneous ciphertexts on the performance of an encryption system was investigated in IEEE Transactions onInformation Forensics and Security (April 2013). For one data encryption system, the probability of receiving an erroneous ciphertext is assumed to be$\mathbf{\beta }$, where$\mathbf{0}\mathbf{<}\mathbf{\beta }\mathbf{<}\mathbf{1}$. The researchers showed that if an erroneous ciphertext occurs, the probability of an error in restoring plain text using the decryption system is .5. When no error occurs in the received ciphertext, the probability of an error in restoring plain text using the decryption system is$\mathbf{\alpha \beta }$, where$\mathbf{0}\mathbf{<}\mathbf{\alpha }\mathbf{<}\mathbf{1}$. Use this information to give an expression for the probability of an error in restoring plain text using thedecryption system.

Short answer

The require result is$\beta (0.5+\alpha (1-\beta )$

Step-by-step solution

Given information and definitions 

The formula for probability$\mathbf{P}\mathbf{=}\frac{\mathbf{favourableoutcomes}}{\mathbf{totaloutcomes}}$

The general rule of compliment is$P\left(A^c\right)=1-P\left(A\right)$

The general rule of multiplication.$\mathbf{P}\mathbf{(}\mathbf{A}\cap \mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{)}\mathbf{\times }\mathbf{P}\left(\frac{\mathbf{B}}{\mathbf{A}}\right)$

$\beta =$probability of receiving an erroneous ciphertext and$0<\alpha <1$

Apply the above formula for result

$\begin{array}{rcl}P\text{(}A\text{)}& =& \alpha \\ P\left(\frac{B}{A}\right)& =& 0.5\\ P\left(\frac{B}{A^c}\right)& =& \alpha \beta \\ P\text{(}{A}^c\text{)}& =& 1-\beta \\ & & \end{array}$

Now, use multiplication rule, then

$$\begin{gathered} P\left(A \text{and} B\right)=\beta \times 0.5 \\ P\left(A^c \text{and} B\right)=\alpha \beta \left(1-\beta \right) \end{gathered}$$

Since is not possible to obtain a single ciphertext that is both errorless and non-errorless it is appropriate to use the addition rule for disjoint or mutually exclusive events.

$\begin{array}{rcl}P\left(B\right)& =& 0.5\beta +\alpha \beta \left(1-\beta \right)\\ & =& \beta 0.5+\alpha \left(1-\beta \right)\\ & & \end{array}$

Therefore, the require result is$\beta 0.5+\alpha \left(1-\beta \right)$