Statistics Conditional Probability QUESTION: There are two local factories that produce the same type of radios. 
An electronic market sells $41$ radios from factory A  whereas $2$ of them are defective and $26$ radios from factory B whereas $2$ of them are defective.
You choose a random radio of them.
What is the conditional probability that this radio is from factory A if it is given that it is working properly? 
MY ATTEMPT : if $2$ of $41$ radio id defective then the probability to get radio working properly is $\frac{39}{41}$.
 A: So we have three events:
$R$ - radio is working
$A$ - radio is from factory $A$
$B$ - radio is from factory $B$
We have $$R = (A\cap R)\cup (B\cap R)$$ and $$P(R) = P(A)\cdot P(R|A)+P(B)\cdot P(R|B)= {41\over 67}\cdot {39\over 41}+{26\over 67}\cdot {24\over 26}={63\over 67} $$
Now $$P(A|R) ={P(A\cap R)\over P(R)}= {P(A)\cdot P(R|A)\over P(R)}={{39\over 67}\over {63\over 67}}= {13\over 21}$$
A: Using  Bayes’ theorem we have:
$Pr(\text{radio is from factory A|radio is working})=\dfrac{Pr(\text{radio is from factory A}\,\bigcap \, \text{radio is working})}{Pr(\text{radio is working})} = \dfrac{\frac{39}{67}}{\frac{63}{67}} = \dfrac{39}{63} = \dfrac {13}{21}\approx 0.619$
Where
$\bullet$ $Pr(\text{radio is from factory A}\,\bigcap \, \text{radio is working})$ means the probability of the radio being from factory A $\textbf{and}$ it works. So in this case we have $39$ working radios from factory A from a total of $67$ radios ($\frac{39}{67}$).
$\bullet$ $Pr(\text{radio is working}) = \dfrac{63}{67}$ because there are $63$ working radios from a total of $67$.
