A certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively. Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively.
If a smartphone of this model is found out to be detective, what is the probability that this smartphone was manufactured in factory C?
Hint.
Use the Bayes’ theorem.
Solution.
Let $E$ be the event that a smartphone of this model is defective. Let $F_A$ be the event that a smartphone is manufactured by factory A. Similarly for $F_B$ and $F_C$.
By Bayes’s rule, we have
\[P(F_C \mid E) = \frac{P(F_C) \cdot P(E \mid F_C)}{P(E)}.\]
Now, we compute the probabilities on the right hand side.
In the post Overall Fraction of Defective Smartphones of Three Factories, we calculated that
\begin{align*}
P(E) &= P(F_A)\cdot P(E \mid F_A) + P(F_B)\cdot P(E \mid F_B) + P(F_C)\cdot P(E \mid F_C)\\
&= 0.0455.
\end{align*}
(See the post for details.)
Factory C produces $15\%$ of the smartphones, thus $P(F_C)=0.15$.
Also, the defective rate for Factory C is $7\%$. Hence $P(E \mid F_C) = 0.07$.
Inserting these values into the formula above, we get
\[P(F_C \mid E) = \frac{0.15 \cdot 0.07}{0.0455} \approx 0.2308.\]
from Problems in Mathematics https://ift.tt/2qPv8JM
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