Question

An encryption-decryption system consists of three elements: encode, transmit, and decode. A faulty encode occurs in 0.7% of the messages processed, transmission errors occur in 1% of the messages, and a decode error occurs in 0.1% of the messages. Assume the errors are independent. Round your answers to four decimal places (e.g. 98.7654).

(a) What is the probability of a completely defect-free message?

(b) What is the probability of a message that has either an encode or a decode error?

Answer 1

Solution

Let A1, A2 and A3 be respectively the event of fault-free encode, fault-free transmit and fault-free decode,

Then, P(A1) = 1 – 0.007 = 0.993; P(A2) = 1 – 0.01 = 0.99 and P(A3) = 1 – 0.001 = 0.999 .............. (1)

Back-up Theory

For 2 events, A and B,

P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B) …………………...........................................…………(2)

If A and B are independent, P(A ∩ B) = P(A) x P(B) ..………....................................…………………(3)

A^C and B^C are also independent and hence P(A^C ∩ B^C) = P(A^C) x P(B^C) ..…………………………(3a)

(2) and (3) =>

P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A) x P(B) …………………….....................................………(4)

Extending (3), if A1, A2 and A3 are three independent events,

P(A1 ∩ A2 ∩ A3) = P(A1) x P(A2) x P(A3) ...................................................................................... (5)

Now, to work out the solution,

Given, ‘Assume the errors are independent’, (3), (3a) and (5) can be directly applied.

Part (a)

Probability of a completely defect-free message

= P(A1 ∩ A2 ∩ A3)

= 0.993 x 0.99 x 0.999   [vide (5) and (1)]

= 0.9821 Answer 1

Part (b)

Probability of a message that has either an encode or a decode error

P(A1^C or A3^C)

= P(A1^C) + P(A3^C) – {P(A1^C) x P(A3^C)} [vide (4)]

= 0.007 + 0.001 – 0.000007

= 0.007993 Answer 2

DONE