TIFR CSE 2013 | Part A | Question: 6

Question

You are lost in the National park of Kabrastan. The park population consists of tourists and Kabrastanis. Tourists comprise two-thirds of the population the park and give a correct answer to requests for directions with probability $\dfrac{3}{4}$. The air of Kabrastan has an amnesaic quality, however, and so the answers to repeated questions to tourists are independent, even if the question and the person are the same. If you ask a Kabrastani for directions, the answer is always wrong.

Suppose you ask a randomly chosen passer-by whether the exit from the park is East or West. The answer is East. You then ask the same person again, and the reply is again East. What is the probability of East being correct?

• A. $\left(\dfrac{1}{4}\right)$
• B. $\left(\dfrac{1}{3}\right)$
• C. $\left(\dfrac{1}{2}\right)$
• D. $\left(\dfrac{2}{3}\right)$
• E. $\left(\dfrac{3}{4}\right)$

Comments

Answer would be 1/2 . good read to solve these type of questions     Link

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CONFUSING

BAYES THEOREM IS MORE DIRECT

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Although I understod the question and solved it completely, But I was thinking, Why haven’t we taken the case in denominator where-

1. the tourist first speaks correctly and then he speaks it wrong
2. OR firstly when the tourist speaks wrongly and then corrects it. 

I means in denominator if all the cases are taken into consideration, the probablity will approximate 3/8

@Ayush Upadhyaya @srestha  @Warrior

Answer 1

upvote_______________________

Answer 2

In this Qn, we need to find out Prob of the correct answer.

1. The correct answer is given by only Tourists.

P(Correct ans) =P( Correct ans by tourist ∩ Tourist)  +  P( Correct ans by kabrastani∩ Kabrastani)

P( Correct ans by kabrastani ∩ Kabrastani) = 0 ,bcoz kabrastani always give wrong ans.

So,

P(Correct ans) = P( Correct ans by tourist ∩ Tourist) = P(Correct ans by tourist) .P(Tourist) = 3/4 .2/3 =1/2

The correct answer is (c) 1/2

Comments

what happens to second time asking, shouldn't this be multiplied by 2?

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only tourist can say correct answer. so prob choosing tourist is 2/3 then he (correct wrong) or (correct correct) or (wrong correct) or (wrong wrong) . as here both times is east so possiblities can be (correct correct) or (wrong wrong)  but finally whatever he said it should be correct. so only possiblility we left is (correct correct)

 

so prob he saying correct is 3/4.

so tourish saying (correct correct) is 2/3 * 3/4 * 3/4 = 3/8.........where iam doing wrong

Answer 3

Let $E_1$ be the event when randomly selected person said the direction is east for the first time, and $E_2$ be the event when same person said direction is east the second time. We need to find the probability that indeed direction is east given $E_1$ and $E_2$.

$$\begin{align}
P(E \ | \ E_1, E_2) &= \dfrac{P(E, E_1, E_2)}{P(E_1, E_2)} \\
&= \dfrac{P(E, E_1, E_2)}{P(E, E_1, E_2) + P(W, E_1, E_2)} \\
&= \dfrac{P(E, E_1, E_2, T) + P(E, E_1, E_2, K)}{P(E, E_1, E_2) + P(W, E_1, E_2)}\\
&= \dfrac{P(E, E_1, E_2, T) + P(E, E_1, E_2, K)}{P(E, E_1, E_2, T) + P(E, E_1, E_2, K) + P(W, E_1, E_2, T) + P(W, E_1, E_2, K)}
\end{align}$$

Where $E$ is the event that direction is indeed east, similarly $W$ is the event that direction is west. $T$ be the event that the randomly selected person is tourist, similarly $K$ be the event that randomly selected person is kabrastani.

$$\begin{align}
P(E \ | \ E_1, E_2) &= \dfrac{\frac{2}{3}\frac{3}{4}\frac{3}{4} + \frac{1}{3}.0.0}{\frac{2}{3}\frac{3}{4}\frac{3}{4} + \frac{1}{3}.0.0 + \frac{2}{3}\frac{1}{4}\frac{1}{4} + \frac{1}{3}.1.1}  = \dfrac{1}{2}\\\end{align}$$

$\textbf{Option (C) is correct}$