%% probability %%

Bayes Rule

Tags

PAS

Cegep/2

Word count

105 words

Reading time

1 minute

$$P(A|B)=\frac{P(A)P(B|A)}{P(B)}$$

Examples

Three machines, A, B and C, are responsible for 50, 30 and 20 percent of the total production of a given item, respectively. The machines have defective rates of 2, 4 and 6 percent, respectively. An item is randomly selected from the lot. If it is defective, find the probability that it was produced by machine A.

$$\begin{array}{rl}P(D)& =P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)\\ & =0.02\cdot 0.5+0.04\cdot 0.3+0.06\cdot 0.2\\ & =0.034\end{array}$$$$\begin{array}{rl}P(A|D)& =\frac{P(A)P(D|A)}{P(D)}\\ & =\frac{0.5\cdot 0.02}{0.034}\\ & \approx 0.2941\end{array}$$

Contributors

Ajitani Hifumi

Changelog

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