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Home  >>  CBSE XII  >>  Math  >>  Probability
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A factory produces bulbs,The probability that any one bulb is defective is $\Large \frac{1}{50}$ and they are packed in boxes of 10.From a single box,find the probability that\begin{array}{1 1}(i)\;none\;of\;the\;bulbs\;is \;defective.\\(ii)\;exactly\;two\;bulbs\;are\;defective\\(iii)\;more\;than\;8\;bulbs\;work\;properly\end{array}

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Toolbox:
  • In a factory probability of defective bulbs is
  • P(defective)=\(\large\frac{1}{50}\)
  • p=\(\large\frac{1}{50}\) \( q=1-P\) =\(\large\frac{49}{50}\)
  • A box contains 10 bulbs
  • n=10
  • P(x=r)=\(^n\)c\(_r\) p\(^r\)q\(^{n-r}\)
P(none defective)=\(p(x=0)\)
=\(^{10}\)c\(_0\)(\(\frac{1}{50}\))\(^0\)(\(\frac{49}{50}\))\(^{10}\)
=(\(\frac{49}{50}\))\(^{10}\)
p(exactly 2defective)=P(x=2)
=\(^{10}\)c\(_2\)(\(\frac{1}{50}\))\(^2\)(\(\frac{49}{50}\))\(^8\)
=45\(\left(\frac{49^8}{50^{10}}\right)\)
P(more then 8 working property)
=P(less than 2 not working)
=P(less than 2 defeective)
=P(x=0)+P(x=1)
=\(^{10}\)c\(_0\)(\(\frac{1}{50}\))\(^0\)(\(\frac{49}{50}\))\(^{10}\)+\(^{10}\)c\(_1\)(\(\frac{1}{50}\))\(^1\)(\(\frac{49}{50}\))\(^9\)
=(\(\frac{49}{50}\))\(^9\)+10x\(\left(\frac{49^9}{50^{10}}\right)\)
=\(\left(\frac{49^9}{50^{10}}\right)\)[49+10]
=59\(\left(\frac{49^9}{50^{10}}\right)\)

 

answered Feb 28, 2013 by poojasapani_1
edited Mar 15, 2013 by poojasapani_1
 

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