Question

37. To improve the reliability of the channel described in the last example, we repeat each digit in the message three times. What is the probability that 111 was sent given that (a) we received 101? (b) We received 000?

Answer 1

a) The probability any digit matches what is sent is 0.9. If we received 101

one digit (the 0) is wrong: 0.9 * 0.1 * 0.9 = 0.081
two digits (the 1s) are wrong: 0.1 * 0.9 * 0.1 = 0.009.

We know that we received 101, and one of these are occurred,

So that the probability 111 was sent is
= 0.081 / (0.081 + 0.009)
= 0.081 / 0.090
= 0.9

b) There are two cases in which we could have received 000:

000 was sent:
= 0.9*0.9*0.9
= 0.729

111 was sent
= 0.1*0.1*0.1
= 0.001

The probability that 111 was sent is therefore
= 0.001 / (0.729 + 0.001)
= 0.001 / 0.730
= 1/730