# Problem about conditional probability with binary digits

Hey I want to check if my solutions are right:

Binary digits are being transmitted over a noisy channel. The probability of a $$'0'$$ or $$'1'$$ being transmitted is each $$0.5$$. The probability that the received digit corresponds to the transmitted digit is $$0.9$$.

(a) What is the probability that a $$1$$ was transmitted if a $$1$$ was received?

(b) As a safety measure, the digit to be transmitted is sent three times (i.e. 111 or 000). What is the probability that a '1' was intended to be transmitted if 111 was received? What is the probability that a '1' was intended to be transmitted if it is known that the sum of the three received digits is equal to 2?

Solution:

I used Bayes' Theorem to find the probability that a $$1$$ was transmitted given that a $$1$$ was received.

$$P(A|B) = P(B|A) * P(A) / P(B)$$

Where $$A$$ represents the event of a $$1$$ being transmitted and $$B$$ represents the event of a $$1$$ being received.

We know that $$P(A) = 0.5$$ (since the probability of a 1 being transmitted is 0.5) and $$P(B|A) = 0.9$$ (since the probability of a 1 being received given that a 1 was transmitted is 0.9).

To find $$P(B)$$, we can use the law of total probability:

$$P(B) = P(B|A) * P(A) + P(B|A^c) * P(A^c)$$

where $$A^cA$$ represents the event of a $$0$$ being transmitted.

Since the probability of a $$0$$ being transmitted is also $$0.5$$, we have:

$$P(B) = 0.9 * 0.5 + 0.1 * 0.5 = 0.5$$

Therefore, using Bayes' Theorem, we have:

$$P(A|B) = P(B|A) * P(A) / P(B) = 0.9 * 0.5 / 0.5 = 0.9$$

So the probability that a 1 was transmitted given that a 1 was received is 0.9.

For b) I am having some problems:

We are searching M= "111" received and "111" was sent

So the probability is $$P(M)=P(B|A)^3$$ Since $$P(M)=P(\text{First Digit: 1 received and 1 Was sent } \cap \text{2nd Digit: 1 received and 1 Was sent }\cap \text{3th Digit: 1 received and 1 Was sent })$$

The three events are independent therefore $$P(M)=P(B|A)^3=\frac{9}{10}^3$$.

For the second part we could have reiceved "011,101,011" So, let G= reiceved "011,101,011" sent "111"

We have $$P(G)=P(A|B)^2*P(A^c|B)=0,9^2*0,1$$

Have I done any mistake? Thanks for the help.

I agree with your solution for the first part, but correct the definition of $$A^c$$(you've typed $$A^cA$$, and I don't have enough reputation to comment). However, I have a different solution for the second part. Please verify and convince yourself which one is correct.

I use $$R$$ for the random variable received and $$S$$ for the random variable sent. The first half of part (b) requires $$P(S = 111|R = 111)$$.

$$p := P(S = 111|R = 111) = \frac{P(R = 111|S = 111)*P(S = 111)}{P(R = 111|S = 111)*P(S = 111) + P(R = 111|S \neq 111)*P(S \neq 111)}$$

$$P(S = 111) = (0.5)^3$$ and $$P(R = 111|S = 111) = (0.9)^3$$ as each intended digit is successfully received, and I assume the sending and reception of a digit is independent of sending or reception of another digit.

Coming to the denominator, the second term in the sum is we receive 111 even though we made a typo while sending. The probability that there is a typo while sending is $$P(S \neq 111) = 1 - P(S = 111) = 1 - (0.5)^3 = P(S = 000) + P(S = 001) + P(S = 010) + P(S = 100) + P(S = 011) + P(S = 101) + P(S = 110).$$

Finally, the conditional probability $$P(R = 111|S \neq 111)$$ is $$P(R = 111|S = 000) + P(R = 111|S = 001) + P(R = 111|S = 010) + P(R = 111|S = 100) + P(R = 111|S = 011) + P(R = 111|S = 101) + P(R = 111|S = 110).$$

$$P(R = 111|S = 000) = (0.1)^3$$ as all three digits are corrupted by the channel/transmitting process.

$$P(R = 111|S = 001) = P(R = 111|S = 010) = P(R = 111|S = 100) = (0.1)^2(0.9)$$ as one digit is correctly transmitted and two are corrupted(again assuming independence).

$$P(R = 111|S = 011) = P(R = 111|S = 101) = P(R = 111|S = 110) = (0.9)^2(0.1)$$ as one digit is corrupted and two are correctly transmitted.

Substituting these values, I get

$$p = \frac{(0.5)^3(0.9)^3}{(0.5)^3(0.9)^3 + [1 - (0.5)^3]*[(0.1)^3 + 3(0.1)^2(0.9) + 3(0.9)^2(0.1)]} = \frac{729}{2626} = 0.2776$$.

Similarly, we can do the second half of part (b).