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Home  >>  CBSE XII  >>  Math  >>  Probability
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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} $

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1 Answer

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  • \(E_1=item\;is\;manufactured\;by\;machin\;A\)
  • \(E_2==item\;is\;manufactured\;by\;machin\;B\)
  • \(E_3=item\;is\;manufactured\;by\;machin\;c\)
  • 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)\)=
\(\huge\frac{\frac{1}{2}\times\frac{2}{100}}{\frac{1}{2}\times\frac{2}{100}+\frac{3}{10}\times\frac{2}{100}+\frac{1}{5}\times\frac{3}{100}}\)
=\(\Large\frac{10}{22}\)
=\(\Large\frac{5}{11}\)

 

answered Mar 2, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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