Confused with probability

Question

In a bit string generated by a random source, the probability of a 1 is 0.6 and the probability of a 0 is 0.4. The bits are transmitted over a noisy communications channel and they are received as sent with probability 0.8, but errors occur with probability 0.2.

1. What is the probability that a 1 was sent given that we received a 1? I was thinking to use bayes theorme but I am not sure how to calculate p(no error | 1sent)

2.To improve the reliability of the channel, we generate a digit and send this digit three times. What is the probability that 111 was sent given that we received 010?

Can someone ples help me! I look forward for reply

Answer 1

Bayes rule is the right idea. Try dealing with the events "receive 0" or "receive 1" instead of "error" and "no error".

$$P(\text{send 1} \mid \text{receive 1}) = \frac{P(\text{receive 1} \mid \text{send 1}) \ P(\text{send 1})}{P(\text{receive 1} \mid \text{send 1}) \ P(\text{send 1}) + P(\text{receive 1} \mid \text{send 0}) \ P(\text{send 0})}$$

The same idea can be used for question two. If you send "111" then each bit has $0.2$ chance of having an error, so the probability of receiving 010 given that 111 was sent is $(0.8)(0.2)^2$. Use Bayes rule to "reverse" the conditioning.

Answer 2

Let's define some variables first:

$A$ is the event that a 1 is sent

$B$ is the event that a 1 is received

Then:

$\operatorname{P}(A) = 0.6$

$\operatorname{P}(B) = \operatorname{P}(\text{no errors} \cap \text{1 was sent}) + \operatorname{P} (\text{error} \cap \text{2 was sent}) = \operatorname{P}(\text{no errors}) \operatorname{P}(\text{1 was sent}) + \operatorname{P} (\text{error}) \operatorname{P} (\text{2 was sent}) = (0.8)(0.6) + (0.2)(0.4) = 0.48 + 0.08 = 0.56$

$\operatorname{P}(B \vert A) = \operatorname{P}(\text{1 received given 1 sent, aka no errors}) = 0.8$

Using Bayes' Rule:

$$\operatorname{P}(A \vert B) = \frac{\operatorname{P}(B \vert A) \operatorname{P}(A)}{\operatorname{P}(B)}= \frac{(0.8)(0.6)}{(0.56)} \approx 0.857$$

And you're done.