Probability
10.0 Illustration for understanding the difference between total probability theorem and baye's theorem
10.0 Illustration for understanding the difference between total probability theorem and baye's theorem
Illustration 39. In a bolt factory, machines $A$, $B$ and $C$ manufacture respectively $25\% $, $35\% $ and $40\% $ of the total bolts. Of their output $5$, $4$ and $2$ percent are respectively defective. A bolt is drawn at random. Find the probability that the bolt is defective.
Solution: Let the events be,
${A_1}$ = bolt is manufactured by machine $A$.
Thus, $$P({A_1}) = 0.25$$
${A_2}$ = bolt is manufactured by machine $B$.
Thus, $$P({A_2}) = 0.35$$
${A_3}$ = bolt is manufactured by machine $C$.
Thus, $$P({A_3}) = 0.40$$
If $M$ is the event of getting a defective bolt, then,
$P(M/{A_1})$ = Probability that the bolt drawn is defective given that it is manufactured by machine $A$
$$P(M/{A_1}) = 0.05$$
Similarly,
$$P(M/{A_2}) = 0.04$$
$$P(M/{A_3}) = 0.02$$
By probability theorem,
$$\begin{equation} \begin{aligned} P(M) = P({A_1})P(M/{A_1}) + P({A_2})P(M/{A_2}) + P({A_3})P(M/{A_3}) \\ P(M) = [0.25 \times 0.05] + [0.35 \times 0.04] + [0.40 \times 0.02] \\ P(M) = 0.0345 \\\end{aligned} \end{equation} $$
Illustration 40. In a bolt factory, machines $A$, $B$ and $C$ manufacture respectively $25\% $, $35\% $ and $40\% $ of the total bolts. Of their output $5$, $4$ and $2$ percent are respectively defective. A bolt is drawn at random. Find the probability that the bolt is made by the machine $B$, given that it is defective.
Solution: Let the events be,
${A_1}$ = bolt is manufactured by machine $A$.
Thus, $$P({A_1}) = 0.25$$
${A_2}$ = bolt is manufactured by machine $B$.
Thus, $$P({A_2}) = 0.35$$
${A_3}$ = bolt is manufactured by machine $C$.
Thus, $$P({A_3}) = 0.40$$
If $M$ is the event of getting a defective bolt, then,
$P(M/{A_1})$ = Probability that the bolt drawn is defective given that it is manufactured by machine $A$
$$P(M/{A_1}) = 0.05$$
Similarly,
$$P(M/{A_2}) = 0.04$$
$$P(M/{A_3}) = 0.02$$
By Bayes' probability theorem,
Probability that the bolt is manufactured by machine $B$, given that the bolt is defective is
$$\begin{equation} \begin{aligned} P({A_2}/M) = {{P({A_2})P(M/{A_2})} \over {P({A_1})P(M/{A_1}) + P({A_2})P(M/{A_2}) + P({A_3})P(M/{A_3})}} \\ P({A_2}/M) = {{0.35 \times 0.04} \over {0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02}} \\ P({A_2}/M) = {{28} \over {69}} \\ P({A_2}/M) = 0.405 \\\end{aligned} \end{equation} $$
$40\%$ of the defective bolts are produced by $B$.
Note:
By using Total probability theorem, we find out that only $0.0345$ of the bolts are defective.
By using Baye's theorem, we are able to understand even more deeper that, about $40\%$ of the defective bolts are by the machine $B$. Hence, it is better to reduce the usage of machine $B$.
