Question

In a factory 4000 units of an item are produced. There are three machines

A, B and C in the factory. B and C produce equal number of items while

machine A produces double items than B. the proportion of defective items

produced by A, B and C are 2%, 1% and 3% respectively. An item is

selected at random from the total production and it is found to be

defective. Find the probability that it comes from machine C.​

Answer 1

$\huge{ \underline{ \overline{ \purple{ \mathfrak{ Answer : }}}}}$

$\sf{Given \: the \: item \: is \: defective, \: probability \: of \: it \: being \: produced \: by \: B = 0.26}$

$\huge{ \underline{ \overline{ \purple{ \mathfrak{ Explanation:}}}}}$

prior information :-

Probability of an item produced by Machine

• $\sf{P(A)=50%=0.5}$

Probability of an item produced by Machine B

• $\sf{P(B)=30%=0.3}$

Probability of an item produced by Machine C

• $\sf{P(C)=20%=0.2}$

On the basis of additional information :-

Given the item produced by A, probability of it being defective

• P(D/A)=3%=0.03

Given the item produced by B, probability of it being defective

• P(D/B)=2%=0.02

Given the item produced by C, probability of it being defective

• P(D/C)=1%=0.01

Given the item is defective, probability of it being produced by B

‎ ‎ ‎ ‎ ‎P(B)×P(D/B)

=‎ ‎ _______________________________

‎ ‎ ‎ ‎ ‎[P(B)×P(D/B)]+[P(A)×P(D/A)]+[P(C)×P(D/C)]

‎ ‎ ‎ ‎ 0.3×0.02

=‎ ‎ ________________________

‎ ‎ ‎ ‎ ‎ ‎ [0.3×0.02]+[0.5×0.03]+[0.2×0.01]

‎ ‎ ‎ ‎ 0.006

=‎ ‎ ________________

‎ ‎ ‎ ‎ 0‎ ‎ .006+0.015+0.002

‎ ‎‎ ‎  0.006

=‎ ‎ ______

‎ ‎ ‎ ‎ 0.023

=0.26

Given the item is defective, probability of it being produced by B = 0.26