There are 4 districts in the land of Oz. At home, the inhabitants of each region wear ties of a special colour, Munchkins (M) wear blue, Scarecrows (S) wear purple, Tin Men (T) wear red and Wizards (W) wear yellow. When visiting the Emerald city however, some inhabitants wear green ties, 25% of Munchkins, 35% of Scarecrows, 45% of Tin Men and 55% of Wizards. As a visitor approaches, the gatekeeper of the Emerald city who knows the tourism rate for the last few years assigns the probabilities as follows:

$P(M)=1/3, P(S)=1/4; P(T)=1/6; P(W)=1/4$

  1. What is the probability that the visitor will wear a green tie?

  2. Given that a visitor is wearing a green tie, calculate the probability that the visitor is a Munchkin.

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum May 4 '15 at 8:27
  • $\begingroup$ The thing is I don't even know where to start. So if any does know how to solve it, could you explain it to me please, thank you $\endgroup$ – Jenelle Petrié May 4 '15 at 9:19

i) Let G denote green ties. So $$P(G) = P(M \cap G) + P(S\cap G) + P(T\cap G) + P(W\cap G) \\ = (1/3) (.25) + (1/4) (.35) + (1/6)(.45)+ (1/4)(.55) \\= 23/60 $$.

ii) This is an application of Bayes' Theorem. $$P(M|G) = \dfrac{P(M\cap G)}{P(G)} = \dfrac{(1/3)(.25)}{(23/60) } = 5/23. $$

  • $\begingroup$ ANumosh thank you so much, I really appreciate your help 😃🙌 $\endgroup$ – Jenelle Petrié May 4 '15 at 10:19

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