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# An item is manufactured by three machines A,B and C.Out of the total number of items manufactured during a specified period,50% are manufactured on A,30% on B and 20% on C.2% of the items produced on A and 2% of items produced on B are defective,and 3% of these produced on C are defective.All the items are stored at one godown.One item is drawn at random and is found to be defective.What is the probability that it was manufactured on machine A?

\$\begin{array}{1 1}(A)\;\large\frac{5}{11}\$B)\;\large\frac{6}{11}\$$C)\;\large\frac{3}{11}\\(D)\;\large\frac{9}{11}\end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • \(E_1=item\;is\;manufactured\;by\;machin\;A$ • $E_2==item\;is\;manufactured\;by\;machin\;B$ • $E_3=item\;is\;manufactured\;by\;machin\;c$ • A=item is defective • $P(A/E_1)=P(item\;found\;defective/item\;is\;manufactured\;by\;machin\;A$) • P(E$_2$/A)=$\Large\frac{{P(E_2)}{P(A/E_2)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}$ • $since\; 50%\;of\; item\; are\; manufactured\; by\; A$ $P(E-1)$=$\Large\frac{50}{100}$=$\Large\frac{1}{2}$ = $30%\;of\; item\; are\; manufactured\; by\; B$ $P(E_2)$=$\Large\frac{30}{100}$=$\Large\frac{3}{10}$ = $20%\;of\; item\; are\; manufactured\; by\; c$ $P(E_1)$=$P(one\; in\; thousand \;having\; T\;B$) =$\Large\frac{1}{1000}$ $p(E_3)=1-\Large\frac{20}{100}=\frac{1}{5}$ $since\; 2% \;produced\; by\; A\; machin\; is\; defective$ P(A/E$_1)$=$\Large\frac{2}{100}$ $since\; 2%\; produced\; by\; B \;machin\; is\; defective$ $P(A/E_2)$=$\Large\frac{2}{100}$ $since\; 3%\; produced\; by\; C\; machin\; is\; defective$ $P(A/E_3)$=$\Large\frac{23}{100}$ $P(E_1$A)$$=
$\huge\frac{\frac{1}{2}\times\frac{2}{100}}{\frac{1}{2}\times\frac{2}{100}+\frac{3}{10}\times\frac{2}{100}+\frac{1}{5}\times\frac{3}{100}}$
=$\Large\frac{10}{22}$
=$\Large\frac{5}{11}$

edited Jun 4, 2013