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