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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?

1. $\left(\dfrac{1}{4}\right)$
2. $\left(\dfrac{1}{3}\right)$
3. $\left(\dfrac{1}{2}\right)$
4. $\left(\dfrac{2}{3}\right)$
5. $\left(\dfrac{3}{4}\right)$

edited | 707 views
0

We have to find the probability of East being correct given that we twice get the answer as East from some person.

Let's denote:

Probability of East being correct $=P(E)$

& Probability of West being correct $=P(W)$

Since, only East and West are possible answers we can assume $P(E) = P(W) = \dfrac{1}{2}.$

Now, the passer-by is giving me $2$ times the answer as East. Lets denote the probability for this as $P(E^{''})$

As we need to find the probability of East being correct given that the passer-by gave $2-$ times East as answer, we can denote this as $P(E\mid E^{''})$

Now, as per Baye's Theorem we can break $P(E\mid E^{''})$ as

$P(E\mid E^{''})$ $= \dfrac{P(E \cap E^{''})}{P(E^{''})} \\ =\dfrac{ P(E) P(E^{''} \mid E)}{P(E^{''})}$

Now, $P(E^{''}) = P(E) P(E^{''} \mid E) + P(W) P(E^{''} \mid W)$

as we can get $2-$ times East in two ways

1. It is an East and someone answers $2$ times East
2. It is a West and someone answers $2$ times East

$\therefore P(E\mid E^{''}) =\dfrac{ \dfrac{1}{2} \times P(E^{''} \mid E)}{\dfrac{1}{2} \times P(E^{''} \mid E) + \dfrac{1}{2} \times P(E^{''} \mid W)} \\ \qquad= \dfrac{ P(E^{''} \mid E)}{P(E^{''} \mid E) + P(E^{''} \mid W)}$

Now, $P(E^{''}\mid E) =$  Probability of getting $2-$ times East given that East is correct.

This we can get in $2$ ways -

1.  When we're asking a tourist & she tells $2-$ times East as answer and East is correct. Probability of this $= \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{3}{4}$
2.  When we're asking a Kabrastani & she tells $2-$ times East as answer and East is correct. Probability of this $= \dfrac{1}{3} \times 0 \times 0$ as Kabrastani always lies.

$\therefore P(E^{''} \mid E) = \text{ Case 1 + Case 2} \\ = \left( \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{3}{4} \right) + \left( \dfrac{1}{3} \times 0 \times 0 \right) \\ = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{3}{4}$

$P(E^{''}\mid W) =$Probability of getting 2-times East given that West is correct.

This we can get in $2$ ways -

1. When we're asking a tourist & she tells $2-$ times East as answer and East is wrong. Probability of this $= \dfrac{2}{3} \times \dfrac{1}{4} \times \dfrac{1}{4}$
2. When we're asking a Kabrastani & she tells $2-$ times East as answer and East is wrong. Probability of this $= \dfrac{1}{3} \times 1 \times 1$

$P(E^{''}|W)$ $= \text{ Case 1 + Case 2} \\ = \left( \dfrac{2}{3} \times \dfrac{1}{4} \times \dfrac{1}{4} \right)+ \left( \dfrac{1}{3} \times 1 \times 1\right)$

$\therefore P(E\mid E^{''})$ $= \dfrac{ P(E^{''} \mid E)}{P(E^{''} \mid E) + P(E^{''} \mid W)}$
$\qquad \qquad = \dfrac{ \left(\dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{3}{4} \right)}{ \left(\dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{3}{4}\right) + \left( \dfrac{2}{3} \times \dfrac{1}{4} \times \dfrac{1}{4} \right) + \left( \dfrac{1}{3} \times 1 \times 1\right) }$
$\qquad \qquad = \dfrac{\dfrac{3}{8}}{ \left(\dfrac{3}{8} \right) + \left(\dfrac{1}{24} \right) + \left(\dfrac{1}{3} \right) }$
$\qquad \qquad = \dfrac{\dfrac{3}{8}}{\dfrac{\left( 9+1+8 \right)}{24}}$
$\qquad \qquad = \dfrac{ \left(\dfrac{3}{8}\right) }{ \left(\dfrac{18}{24} \right)}$
$\qquad \qquad = \left( \dfrac{3}{8} \right) \times \left( \dfrac{24}{18} \right)$
$\qquad \qquad = \dfrac{1}{2}$

Correct Answer: $C$

by Boss (17.7k points)
edited
0

I tried to solve this question using tree method. Given below is my solution. I didnt consider the scenario of asking second time, still I am getting correct answer. Please tell what I am missing.

It's a Baye's theorem question

The population of tourists is $\frac{2}{3}$ and that of kabristanis is $\frac{1}{3}$.

So, the probability that from a random group of people, a tourist is chosen is 2/3 and kabristani is chosen is 1/3.

Now it is given that

the answers to repeated questions to tourists are independent, even if the question and the person are the same.

That simply means tourists don't change their answer according to the situation and if one particular question is asked to them any number of times their reply will be same with true being with a probability of 3/4 and incorrect with probability of 1/4.

If you ask a Kabrastani for directions, the answer is always wrong.

Means, it doesn't matter whether how many times you ask a kabrastani for directions, he will always lie.

Now, with below Image I model the my Baye's theorem Now the probability of getting the wrong answer will be the sum of all the blue circles

By law of total probability it becomes=$\frac{2}{3}*\frac{1}{4} + \frac{1}{3}*1=\frac{1}{2}$

So, the probability that answer is correct=$\frac{1}{2}$

by Boss (27.3k points)
Ans should be  $\dfrac{1}{2}$

Tourists are $\dfrac{2}{3}$ of population of the park

Among them probability of correct answer is $\dfrac{3}{4}$

so, $\dfrac{2}{3}\times \dfrac{3}{4}=\dfrac{1}{2}.$
by Veteran (117k points)
edited
0
my doubt is why here bays theorem is not applied...intially i thought about bays theorem..??
0
You can solve this que by using Conditional property also.I think the best way to understand this Qn by using CP.
0
why we are not considering the second case of asking again the same person about the direction in above solution????
+1

@srestha Isn't there any use of this?

You then ask the same person again, and the reply is again East.

Let

E = East is the correct answer

S = Same person answers East 2 times.

We are asked to find p(E|S) :

= $P(EΠS) / P(S)$

= $P($East is correct and same person is telling east 2 times $) \div P($ same person saying east 2 times $)$

= $P($ Same person is telling correct answer 2 times $) \div P($ Same person saying east 2 times, no matter if East is correct or not$)$

$=\frac{2/3 * 3/4* 3/4 + 1/3* 0*0}{2/3*3/4*3/4 + 2/3*1/4*1/4 + 1/3 * 1*1+1/3 *0*0}$
$= \frac{1}{1+1/9+1/2*4/3*4/3}$
$=\frac{1}{2}$

where 0 is showing that the kabrastani person will tell correct answer with probabilty 0.
by Active (2.7k points)
edited by
+1
Shouldn't we also add a factor of probability of any person selecting 'East'  as an answer , which is 1/2 to each term ?
0
Yes.

But that will be added to each term and ultimately cancelled ;)
+1 vote upvote_______________________

by Active (4.8k points)

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

by Loyal (7.8k points)
0
what happens to second time asking, shouldn't this be multiplied by 2?
0
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