Let E1=the item is manufactured by the operator A, E2=the item is manufactured by the operator B, E3=the term is manufactured by the operator C and A=the item is defective
Now P(E1)=\(\frac{50}{100}\),P(E2)=\(\frac{30}{100}\),P(E3)=\(\frac{20}{100}\)
Now P(A|E1)=P(item drawn is manufactured by operator A)=\(\frac{1}{100}\)
Similarly,P(A|E2)=\(\frac{5}{100}\) and P(A|E3)=\(\frac{7}{100}\)
Now required probability =probability that the item is manufactured by operator A given that the item drawn is defective
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}\)
=\(\frac{50}{100}\)\(×\frac{1}{100}.\frac{50}{100}\)×\(\frac{1}{100}\)+\(\frac{30}{100}\)×\(\frac{5}{100}\)+\(\frac{20}{100}\)×\(\frac{7}{100}\)=\(\frac{50}{50}\)+150+140=\(\frac{5}{34}\)
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Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.