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GO Classes DPP | GATE DA | Probability | Conditional Probability

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A medical laboratory buys testing kits from two manufacturers, $A$ and $B$. Manufacturer $A$ supplies $65\%$ of the kits, and manufacturer $B$ supplies $35\%$ of the kits. All kits are inspected for accuracy. A kit that passes inspection is called certified. Of manufacturer $A$'s kits, $92\%$ are certified. Of manufacturer $B$'s kits, $80\%$ are certified.

The probability that a randomly chosen testing kit, given that it is certified, was made by $B$ is

  1. $0.319$
     
  2. $0.350$
     
  3. $0.427$
     
  4. $0.800$

2 Answers

4 4 votes
Let $C$ be the event that a kit is certified.

We want $P(B \mid C).$

By Bayes' theorem,
$P(B \mid C)=\frac{P(B \cap C)}{P(C)}.$

Now,
$P(B \cap C)=P(B)P(C \mid B)=0.35 \cdot 0.80=0.28.$

Also,
$P(C)=P(A)P(C \mid A)+P(B)P(C \mid B).$

$\Rightarrow P(C)=0.65 \cdot 0.92 + 0.35 \cdot 0.80=0.598+0.28=0.878.$

$\therefore P(B \mid C)=\frac{0.28}{0.878}\approx 0.319.$

Answer: $\boxed{\text{A. 0.319}}$
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