Probability problem using bayesian rule?

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.

(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 ?

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

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

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

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

Please use bayesian rule to solve the problem. Also show what are your random variables.

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Here are some of my attempts.

(a) P(the answer is correct) = P(the answer is correct | ask a visitor)*P(ask a visitor) + P(the answer is correct | ask a Bandrika)*P(ask a Bandrika) = 3/4*2/3 + 0*1/3 = 1/2;

(b) P(ask the same people twice and the answer is correct | ask a visitor) = 3/4*3/4; P(ask the same people twice and the answer is correct) = P(ask the same people twice and the answer is correct | ask a visitor)*P(ask a visitor) = 9/16*2/3=3/8. :( this is not correct...

(c) because (b) is incorrect, so the behind problem I cannot answer.

Answer 1

For a Bayesian perspective, you need to properly select your hypothesis and your evidence. In this case, your hypotheses are simple:

$H_1 =$ East is correct, $H_2 =$ East is not correct.

The evidence you're given at each stage is of the form

$E_n =$ you get the answer East $n$ times in a row from the same person.

So, for (b), you want $P(H_1|E_2)$ and the general Bayesian formula is

$P(H_1|E_2)={P(E_2|H_1)P(H_1)\over P(E_2|H_1)P(H_1)+P(E_2|H_2)P(H_2)}$.

In the absence of any other info, we assume East and West are equally likely to be the right answer, so $P(H_1)=P(H_2)={1\over 2}$, which cancels those out across the fraction leaving you with

[1] $P(H_1|E_2)={P(E_2|H_1)\over P(E_2|H_1)+P(E_2|H_2)}$.

Now, $P(E_2|H_1)$ = the probability of East twice from the same person given that East is correct = (prob of picking a tourist)(prob of 2 right ans) + (prob of picking a local)(prob of 2 right ans) = $({2\over 3})({3\over 4})({3\over 4}) + ({1\over 3})(0)(0)$

and $P(E_2|H_2)$ = the probability of East twice from the same person given that East is not correct = (prob of picking a tourist)(prob of 2 wrong ans) + (prob of picking a local)(prob of 2 wrong ans) = $({2\over 3})({1\over 4})({1\over 4}) + ({1\over 3})(1)(1)$. So we get

$P(H_1|E_2)={P(E_2|H_1)\over P(E_2|H_1)+P(E_2|H_2)}={{({2\over 3})({3\over 4})({3\over 4}) + ({1\over 3})(0)(0)}\over {({2\over 3})({3\over 4})({3\over 4}) + ({1\over 3})(0)(0)+({2\over 3})({1\over 4})({1\over 4}) + ({1\over 3})(1)(1)}}={1\over 2}$

(c) and (d) just increase the repeated "answer" probabilities in each term by another factor, and once you've written out the equation for case (d), you should be able to see where to change the "fourth question" probablities from "right" to "wrong" across the board to answer (e).

To clarify a bit, the answers you get to your questions under $E_2$ are East, East, and the $H_1$ in $P(E_2|H_1)$ means for these terms East is correct, so the terms of $P(E_2|H_1)$ are $({2\over 3})({3\over 4})({3\over 4}) + ({1\over 3})(0)(0)$ and correspond to

(tourist)(East)(East)+(local)(East)(East)

So, in part (e), let's call it $E_5$, the answer sequence is (East)(East)(East)(West), so

$P(E_5|H_1)=({2\over 3})({3\over 4})({3\over 4})({3\over 4})({1\over 4}) + ({1\over 3})(0)(0)(0)(1)$

Which you can read as "The probability of (East)(East)(East)(West) given that East is correct = (prob of choosing tourist)(tourist correct)(tourist correct)(tourist correct)(tourist wrong) + (prob of choosing local)(local correct)(local correct)(local correct)(local wrong).

From that you should be able to construct $P(E_5|H_2)$, and use equation [1] above to compute $P(H_1|E_5)$, which is the answer to (e).

Answer 2

Your mistake was in assuming that getting the same answer twice did not change the chance that you asked a visitor. Visitors will only answer the same way twice $5/16$ of the time, while locals answer the same way $100\%$ of the time.

What you need to do is calculate the probability that the answer is correct, given that the answer is "East". Given that the probability you asked a tourist changes based on how often you got the same answer, the best way of doing this is to make a table of all possible outcomes and then calculate the fraction of the time that $n$ "East" answers occur when the correct answer is "East".

a) Probabilities for asking once and getting "East": $p_1 = $ probability of type ("Tourist" or "Local"), $p_2$ = probability of actual answer (unknown, so always assume $50\%$), $p_3$ = probability of "East" given type and actual: $$ \begin{matrix} \text{Actual:} & \text{East}&\text{West}\\ \text{Type: Tourist} & {2\over 3} \times {1\over 2} \times {3\over 4} = {1\over 4} & {2\over 3} \times {1\over 2} \times {1\over 4} = {1\over 12}\\ \text{Type: Local} & {1\over 3} \times {1\over 2} \times {0} = {0} & {1\over 3} \times {1\over 2} \times {1} = {1\over 6}\\ \text{Total:} &{1\over 4} + 0 = {1\over 4} &{1\over 12} + {1\over 6} = {3\over 12} = {1\over 4} \end{matrix} $$

The probability that the answer is correct is $$ {\text{Total(East)}\over \text{Total(East)} + (\text{Total(West)}} $$

In this case, that is clearly $1/2$.

b) Probabilities for asking twice and getting "East" both times: $$ \begin{matrix} \text{Actual:} & \text{East}&\text{West}\\ \text{Type: Tourist} & {2\over 3} \times {1\over 2} \times {9\over 16} = {3\over 16} & {2\over 3} \times {1\over 2} \times {1\over 16} = {1\over 48}\\ \text{Type: Local} & {1\over 3} \times {1\over 2} \times {0} = {0} & {1\over 3} \times {1\over 2} \times {1} = {1\over 6}\\ \text{Total:} &{3\over 16} + 0 = {3\over 16} &{1\over 48} + {1\over 6} = {9\over 48} = {3\over 16} \end{matrix} $$

In this case, the answer is still $1/2$.

Continue in this vein for the other questions. The only values that change are $p_3$.