Three machines I, II, III set up the $30\%$, $30\%$ and $40\%$ respectively of the total of a television. From them the $4\%$, $3\%$ and $2\%$ respectively are defective.
We choose randomly a television and we check if it is defective or not.
a) Calculate the probability the television that we chosen to be defective.
b) Calculate the probability the television to be constructed by the machine I if we know that it is defective.
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I have done the following :
a) Is the probability equal to $\frac{4}{100}\cdot \frac{30}{100}+\frac{3}{100}\cdot \frac{30}{100}+\frac{2}{100}\cdot \frac{40}{100}$ ?
b) We have to use theorem of Bayes, right? Is it $30\%$ (that it is from machone I) divided by the result of (a) ?
Your answer to the first question is correct.
By Bayes' Theorem, the probability that a television was constructed by machine I given that it is defective is \begin{align*} \Pr(I \mid D) & = \frac{\Pr(I \cap D)}{\Pr(D)}\\ & = \frac{\Pr(D \mid I)\Pr(I)}{\Pr(D \mid I)\Pr(I) + \Pr(D \mid II)\Pr(II) + \Pr(D \mid III)\Pr(III)}\\ & = \frac{\dfrac{4}{100} \cdot \dfrac{30}{100}}{\dfrac{4}{100} \cdot \dfrac{30}{100} + \dfrac{3}{100} \cdot \dfrac{30}{100} + \dfrac{2}{100} \cdot \dfrac{40}{100}} \end{align*}