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Q. A batch of fifty radio sets was purchased from three different companies A, B and C. Eighteen of them were manufactured by A, twenty of them by B and the rest were manufactured by C.
The companies $A$ and $C$ produce excellent quality radio sets with probability equal to 0.9 ; $B$ produces the same with the probability equal to 0.6 .
If the probability of the event that the excellent quality radio set chosen at random is manufactured by the company $B$ is expressed as $\frac{p}{q}$, find the least value of $(p+q)$.

Probability - Part 2

Solution:

Let A, B, C be the events that the radio set chosen, is manufactured by the companies A, B
and C respectively. Let E be the event that the radio set is of excellent quality
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$\text { Thus } P(A)=\frac{18}{50} ; P(B)=\frac{20}{50} ; P(C)=\frac{12}{50} $
$ P(E / A)=0.9 ; P(E / B)=0.6 ; P(E / C)=0.9$
$\therefore P(E)=P(A) P(E / A)+P(B) P(E / B)+P(C) P(E / C)=\frac{39}{50}$
required probability $=P(B / E)=\frac{P(B) P(E / B)}{P(E)}=\frac{4}{13}$