Answer to Question #217968 in Statistics and Probability for Kojo

Question

  11.   (a)       State and prove Bayes’ theorem.

(b)       In a bolt manufacturing factory, machines and  produce 25%, 30% and 45% of the total outputs, respectively. Of their outputs, 7%, 6% and 4% are defective bolt, respectively.

(i)                What is the probability that a bolt drawn at random from production will be defective?

(ii)             If a bolt drawn at random from production is found to be defective, what is the probability that it was manufactured by machine ?

 

Answer 1

Solution:

(a):

Bayes' Theorem describes the probability of occurrence of an event related to any condition. It is also considered for the case of conditional probability.

Statement: Let $E_{1}, E_{2}, \ldots \ldots, E_{n}$ be a set of events associated with a sample space $\mathrm{S}$ , where all the events $E_{1}, E_{2}, \ldots \ldots, E_{n}$ have non zero probability of occurrence and they form a partition of S. Let A be any event associated with S, then according to Bayes' theorem, $P\left(E_{i} \mid A\right)=\dfrac{P\left(E_{i}\right) P\left(A \mid E_{i}\right)}{\sum_{k=1}^{n} P\left(E_{k}\right) P\left(A \mid E_{k}\right)}, k=1,2,3, \ldots, n$

Proof: According to conditional probability formula, $P\left(E_{i} \mid A\right)=\frac{P\left(E_{i} \cap A\right)}{P(A)}$ ...(1)

Using multiplication rule of probability, $P\left(E_{i} \cap A\right)=P\left(E_{i}\right) P\left(A \mid E_{i}\right)$ ...(2)

Using total probability theorem, $P(A)=\sum_{k=1}^{n} P\left(E_{k}\right) P\left(A \mid E_{k}\right)$ ...(3)

Putting the values from equation (2) and (3) in equation (1), we get $P\left(E_{i} \mid A\right)=\dfrac{P\left(E_{i}\right) P\left(A \mid E_{i}\right)}{\sum_{k=1}^{n} P\left(E_{k}\right) P\left(A \mid E_{k}\right)}, k=1,2,3, \ldots, n$

Hence, proved.

(b):

Let

A: bolt produced by machine A,

B: bolt produced by machine B,

C: bolt produced by machine C,

D: bolt produced is defective.

$P(A)=0.25,P(B)=0.30,P(C)=0.45$

$P(D|A)=0.07,P(D|B)=0.06,P(D|C)=0.04$

(i): $P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)$

$=0.25\times0.07+0.30\times 0.06+0.45\times 0.04 \\=0.0535$

(ii): Our question is incomplete. We can assume it as follows:

If a bolt drawn at random from production is found to be defective, what is the probability that it was manufactured by machine B?

$P(B|D)=\dfrac{P(B).P(D|B)}{P(A).P(D|A)+P(B).P(D|B)+P(C).P(D|C)}$

$=\dfrac{0.30\times 0.06}{0.25\times 0.07+0.30\times 0.06+0.45\times 0.04}$

$=\dfrac{36}{107}$