I was solving this problem -

In a city, half of the days have some rain. The weather forecaster is correct 2/3 of the time, i.e., the probability that it rains, given that the forecaster has predicted rain, and the probability that it does not rain, given that she has predicted that it won't rain, are both equal to 2/3. When rain is forecast, Mr.X takes his umbrella. When rain is not forecast, he takes it with probability 1/3. Find (a) the probability that Mr. X has no umbrella, given that it rains. (b) the probability that he brings his umbrella, given that it doesn't rain.

So Far what I have done - 1. Probability Model -

A - {It rains}

B - {Forcaster says it rains}

C - {Mr. X carries umbrella}

Now Given - $$P(A) = 1/2; P(A/B) = 2/3; P(A'/B') = 2/3; P(C/B) = 1; P(C'/B) = 1/3$$ Now P(A'/B') = $(1-P(A \cup B))/P(B') = 2/3$ $$\Rightarrow P(A\cup B) = 1/3+2/3P(B)$$

Also from P(A/B) I get $P(A\cap B) = 2/3P(B)$

So from here I calculated $P(B) = 1/2$ and using this and P(C/B), P(C'/B) I can find P(C) as $$P(C) = 2/3$$ I will apply the sequential tree as - Event1 - Weather forecast Event2 - Mr. X carries umbrella (or not depending upon forecast) Event3 - It rains (or not depending upon forecast)

Probability Model

Hence from this the Probabilities for 1st - 1/9 and 2nd should be 5/18 But I am bit uncertain here. can someone please correct me if I have made any mistake.


  • $\begingroup$ Any comments??.. $\endgroup$ – codeomnitrix Nov 4 '14 at 8:57

(a) Perhaps you calculated $P(C^{'} \cap A)$ instead of $P(C^{'} \mid A)$.

Using the probability tree (event $C^{'} \cap A$ has branch $6$ only):

$$P(C^{'} \cap A) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{1}{9}.$$

$$\therefore P(C^{'} \mid A) = \dfrac{P(C^{'} \cap A)}{P(A)} = \frac{1/9}{1/2} = \frac{2}{9}.$$

(b) Similarly with this one. Using the probability tree (event $C \cap A^{'}$ has branches $2$ and $3$):

$$P(C \cap A^{'}) = \frac{1}{2}\cdot 1 \cdot\frac{1}{3} \;+\; \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{2}{3} = \frac{5}{18}.$$

$$\therefore P(C \mid A^{'}) = \dfrac{P(C \cap A^{'})}{P(A^{'})} = \frac{5/18}{1/2} = \frac{5}{9}.$$

  • $\begingroup$ yeah I realized it, forget to update the answer....thanks Mick\ $\endgroup$ – codeomnitrix Nov 5 '14 at 15:11

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