Find the probability of binary bit when sent to channel

Question

I have a question about Bayesian rule. My question is

For a certain binary communication channel, the prob. that any bit sent is a $0$ is $0.49$. An error occurs with probability $0.08$ given that a $0$ was sent, and with probability $0.01$ given that a $1$ was sent. Find the probability that when a $1$ is received, a $1$ was actually sent?

Could you help me solve above problem? Thanks

Answer 1

Let $S$ be the bit sent and $R$ be the bit received. The task is to find $P(S=1\ |\ R=1)$.

First, we need to know the probability of $P(R=1)$. We know that $P(S = 0) = 0.49$, and thus $P(S = 1) = 1 - 0.49 = 0.51$. We can use them to compute the probability of receiving $1$:

$P(R=1) = P(R=1|S=1) \times P(S=1) + P(R=1|S=0) \times P(S=0)$

$= (1-0.01) \times 0.51 + 0.08 \times 0.49 = 0.5441$.

Now all you have to do is to apply Bayes' rule:

$P(S=1|R=1) = \frac{P(R=1|S=1)P(S=1)}{P(R=1)} = \frac{(1-0.01) \times 0.51}{0.5441} = \frac{0,5049}{0.5441} \approx 92.8\%$.

EDIT: Thanks to Barak Manos for correcting a miscalculation.

Answer 2

Let A denote the event of transmitting a "$1$".

Let B denote the event of receiving a "$1$".

Then $P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{(1-0.49)(1-0.01)}{(1-0.49)(1-0.01)+0.49\cdot0.08}=92.8\%$.