# Probability of signal malfunctioning the same way

I'm writing a paper for computer networks and came across the problem of calculating the probability that a certain event happens:

Problem:

I have two wires, each transmitting one bit of information at a time. Each wire can malfunction and send the wrong bit with the probability of p. if the two wires transmit 32 times, what are the chances that the two wires malfunctioned the same way (eg, a malfunction occurs on both wires at the same time)?

Thinking Process

I was thinking:

Take the first transmission, let X be whether a malfunction occured for wire 1, and Y be weather it happened for wire 2. Then, X and Y follow a bernoulli rv.

P(malfunction happened for both wires for the first bit) = p^2

Now im stuck, as how would I take into account all possibilities for this? I was thinking a binobial rv, but binomials are order independent when this is order dependent.

I was also thinking:

P(1 malfunction happened in the 2 wires) * 32 + P(2 malfunctions happened * 32C2) etc but I'm again, unsure of this as well

Thank you humbly for your help! :)

You're on the right track.

Let's say $X_i$ is an indicator random variable for whether the first wire malfunctions on the $i^{th}$ bit where $X_i = 1$ means a malfunction and $X_i = 0$ means not a malfunction. Let $Y_i$ be the same thing for the second wire. What you ultimately want to calculate is $P(X_1 = Y_1 \wedge X_2 = Y_2 \wedge ... \wedge X_{32} = Y_{32})$.

I'm going to assume that $X_1, ..., X_{32}, Y_1, ..., Y_{32} \sim Ber(p)$ are all IID. The problem doesn't explicitly state this, but it seems to imply it, and we can't find a specific answer without this assumption anyway.

Using that the variables are IID: $P(X_1 = Y_1 \wedge X_2 = Y_2 \wedge ... \wedge X_{32} = Y_{32}) = \prod_{i=1}^{32}P(X_i = Y_i) = P(X_1 = Y_1)^{32}$. Now we need to calculate $P(X_1 = Y_1)$.

$P(X_1 = Y_1)$ is the probability that both wires malfunction or both wires don't malfunction in the first bit. We can calculate this as follows: $P(X_1 = Y_1) = P(X_1 = Y_1 = 1) + P(X_1 = Y_1 = 0) = P(X_1 = 1)P(Y_1 = 1) + P(X_1 = 0)P(Y_1 = 0) = p^2 + (1 - p)^2$.

Thus, putting this back into the above, the probability that the two wires malfunction in the same way over all 32 bits is $(p^2 + (1-p)^2))^{32}$.

• you know they say computer science is hard, I'd say statistics is harder. Thanks! :)
– Wboy
May 28, 2017 at 8:05