Alexsandro said:
Can someone help me with this question ?
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.
Well, arent the Bandrikans nasty!
(a) 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 ?
There is, according to this, a 2/3 chance that the person you asked is a tourist. If that is true, the probability that they answer correctly is 3/4. Probability that the person you ask is a tourist
and answers correctly is (2/3)(3/4)= 1/2. Strictly speaking, we should now add the probability that the person asked is a Bandrikan
and they answer correctly but the probability of that is (1/3)(0)= 0. The probability of getting a correct answer is 1/2.
b)You ask the same person again, and receive the same reply. Show the probability that it is correct is 1/2.
I don't understand this!
Why would we expect the person, whether tourist or Bandrikan, to give a different answer? We are told that the probability that the tourist will answer correctly is 3/4 but I wouldn't assume that two answers in a row are independent! If the tourist believed, rightly or wrongly, that the exit was East and told you that the first time, why wouldn't he/she answer the same the second time? In any case, the answer is exactly what it was in (a): 1/2!
On the other hand, IF we are to assume that the tourist is answering
at random, with probability of being right 3/4 each time, independently of any previous answer (which seems to me a very strange assumption), then I would argue that the probability of a tourist being correct two times in a row is (3/4)(3/4)= 9/16 and the probability of a tourist being wrong two times in a row is (1/4)(1/4)= 1/16. The probability of a tourist being correct two times in a row
given that they gave the same answer two times in a row, is (9/16)/(9/16+ 1/16)= (9/16)(16/10)= 9/10. The probability of asking a tourist
and getting the correct answer two times in a row (under these hypotheses) is (9/10)(2/3)= 3/5, NOT 1/2!
(c) You ask the same person again, and receive the same reply. What is the probability that is correct ?
A third time? I would have to answer 1/2 again because, again, I see no reason for either tourist or Bandrikan to change his/her answer!
IF we are to assume that the tourist is answering
at random, with probability of being right 3/4 each time, independently of any previous answer, then I would argue that there was a (2/3) chance that the person we asked was a tourist and, each time, there was a 3/4 chance of getting a correct answer: The probability that a tourist answers correctly 3 times in a row is (3/4)(3/4)(3/4)= 27/64 and the probability that a tourist answer incorrectly 3 times in a row is (1/4)(1/4)(1/4)= 1/64. The probability that a tourist answers correctly three times in a row
given that they answer the same three times in a row is (27/64)/((27/64)+ (1/64))= (27/64)(64/28)= 27/28. The probability that the person you ask is a tourist
and answer correctly three times in a row would be (2/3)(27/28)= 9/14.
(d) You ask for the fourth time, and receive the answer East. Show that the probability it is correct is 27/70.
This is getting ridiculous! How dumb do they think the tourists are? If I were the tourist and you asked me the same question 4 times in a row, I would punch you in the nose! My answer would still be 1/2, not 27/70!
But: probability that the tourist, answering randomly, gets the question right 4 times in a row is (3/4)
4= 81/256 and the answer that the tourist gets the question wrong 4 times in a row is (1/4)
4= 1/256.
The probability that the tourist gets the question right 4 times in a row
given that they answered the same way 4 times in a row is (81/256)/((81/256)+ (1/256))= (81/256)(256/82)= 81/82. The probability that you asked a tourist and he/she answered correctly 4 times in a row is (2/3)(81/82)= 27/41, not 27/70!
(e) Show that, had the fourth answer been West instead, the probability that that East is nevertheless correct is 9/10.
Oh, I give up!