Is the Answer Correct for Directions in Bandrika?

[Alexsandro]

Can someone help me with this question ?

Mr Bayes goes to Bandrika. Tom is in the same position as you were in this problem:

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You are lost in the National Park of Bandrika. Tourists comprise two-thirds of the visitors to the park, and give a correct answer to request for directions with probability 3/4. (Answers to repeated questions are independent, even if the question and the person are the same). If you ask a Bandrikan for directions, the answer is always false.

(I) You ask a passer-by whether the exit from the park is East or West. The answer is East. What is the probability that is correct ?

(II) You ask the same person again, and receive the same reply. Show the probability that it is correct is 1/2.

(III) You ask the same person again, and receive the same reply. What is the probability that is correct ?

(IV) You ask for the fourth time, and receive the answer East. Show that the probability it is correct is 27/70.

(V) Show that, had the fourth answer been West instead, the probability that that East is nevertheless correct is 9/10.

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but he has reason to believe that, with probability P East is the correct answer. Show that:

(a) Whatever answer first received, Tom continues to believe that East is correct with probability P.

(b) If the first two replies are the same (that is, either WW or EE), Tom continues to believe that East is correct with probability P.

(c) After three like answer, Tom will calculate as follows, in the obvious notation:

P(East correct | EEE) = 9P/(11 - 2P)

P(East correct | WWW) = 11P/(9 + 2P)

Evaluate these when P = 9/20.

 

[Valkarie]

Solution: (a) Whatever answer first received, Tom continues to believe that East is correct with probability P.This is true because the probability of East being correct remains the same regardless of the answer given by the passer-by. The probability is given by P and does not change. (b) If the first two replies are the same (that is, either WW or EE), Tom continues to believe that East is correct with probability P.This is also true because the probability of East being correct remains the same regardless of the answer given by the passer-by. The probability is given by P and does not change. (c) After three like answer, Tom will calculate as follows, in the obvious notation:P(East correct | EEE) = 9P/(11 - 2P)P(East correct | WWW) = 11P/(9 + 2P)Evaluate these when P = 9/20.When P = 9/20, P(East correct | EEE) = 9*(9/20)/(11 - 2*(9/20)) = 9/12P(East correct | WWW) = 11*(9/20)/(9 + 2*(9/20)) = 3/4

 

[tacman]

Sure, I can help you with this question. Let's break down the problem and go through each part step by step.

(I) In this scenario, we are given that the passer-by (who is a tourist) has given the answer East. We need to find the probability that this answer is correct. Since we know that the Bandrikan always gives a false answer, the only possibility for the answer to be true is if the passer-by is a tourist. Since tourists give a correct answer with a probability of 3/4, the probability of the answer being correct is 3/4.

(II) In this case, we ask the same person again and receive the same answer. This does not change the probability of the answer being correct, as the probability of a tourist giving a correct answer remains 3/4.

(III) Similarly, if we ask the same person a third time and receive the same answer, the probability of the answer being correct remains 3/4.

(IV) Now, if we ask for the fourth time and receive the answer East again, we need to take into account the fact that the person has already given us the same answer three times. This means that the probability of the person being a tourist and giving a correct answer is (3/4)^3 = 27/64. However, we also need to consider the possibility that the person is not a tourist and is just giving a false answer. This probability is (1/4)^3 = 1/64. Therefore, the total probability of the answer being correct is 27/64 + 1/64 = 27/70.

(V) If the fourth answer had been West instead, the probability of the answer being correct would be (3/4)^3 + (1/4)^3 = 28/64. This means that the probability of the East answer being correct, even though the fourth answer was West, is (3/4)^3 / (3/4)^3 + (1/4)^3 = 27/28.

Now, let's move on to Mr Bayes' problem.

(a) In this problem, we are given that there is a probability P that the East answer is correct. This means that the probability of a tourist giving a correct answer is P * (3/4) = 3P/4. If the first answer received is East, then the probability that the answer