# Probability of receiving a fixed bit sequence in a binary symmetric channel

What is the probability of receiving 1111 irrespective of what was transmitted?

$0.1^4 + 0.9^4 = 0.6562$

I am not sure about my answer because I think in my solution I am excluding the probability of getting 1111 as a combination of the two (0 or 1). I only assumed that the 1111 is a result of 0 being transmitted $0.1^4$ or 1 $0.9^4$.

Can anyone clarify this point to me?

The probability of receiving 1111 is equal to the probability of transmitting a 1 taken to the fourth power. The probability of transmitting a 1 is equal to the sum of the probability of transmitting a 1 given the input is a 0 plus the probability of transmitting a 1 given the input is a 1. Based on your information that is equal to:
$0.4 \cdot 0.1 + 0.6 \cdot 0.9 = 0.04 + 0.54 = 0.58$
Thus the probability of receiving 1111 is $0.58^4 = 0.1132$