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I need a small confirmation regarding a probability problem:

We estimate that 5% of Americans spent their holidays in Texas, this proportion reaching 40% among Texans. Texans represent 2% of the whole population. We choose randomly an American spending his holidays in Texas. What is the probability that he is a Texan?

My reasoning:

Let P(T): he is a Texan.

P(HT): he is spending is holidays in Texas.

This is Bayes problem.

$ P(T/HT)=\frac { P(T\cap HT) }{ P(HT) } =\frac { 0,02 * 0,4 }{ 0,02 * 0,4 +0,98*0,05 } =0,14 $

Am I correct ?

The weird thing is that 0.14 is not among the proposed answers, which makes me thing that I am wrong.

Thanks in advance.

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3 Answers 3

up vote 4 down vote accepted

according to the question, $$ P(HT) = 0.05\\ P(HT|T) = 0.40\\ P(T) = 0.02 $$hence $$ P(T|HT) = \frac{P(HT|T)P(T)}{P(HT)} = \frac{0.40\times 0.02}{0.05} = 0.16 $$

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Note that you know that 5 % of Americans spend their holidays in Texas, not 5 % of non-Texans. Therefore, $P(\text{HT}) = 0.05$ directly.

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Thanks a lot JiK. You're right. –  XCoder Jun 30 at 13:41

Mookid is correct

The lazy wa to do this is to consider 1000 Americans. 980 are non-Texans. 20 are Texans. 50 holiday in Texas. Of the 20 Texans, 40% or 8 holiday in Texas. Therefore 42 of the 50 texan holidayers are non-Texans. Or 84 out of 100 Texan holidaymakers. So chances of Texan holidaymaker being Texan is 16%.

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