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Home  >>  CBSE XII  >>  Math  >>  Probability
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If X is the number of tails in three tosses of a coin ,determine the standard deviation of X.

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Toolbox:
  • In a toss of 3 coin .X denotes the number of tails
  • X can take value X=0(notail);(X=1)(1 tail)
  • X=2(2tails);X=3(3tails)
  • \(x_i\)=i-1,2,3,
  • \(P_i\)=are there correrpending proabilities
  • =\(\sigma\;\Large\;\sqrt{\sum_iP_i\;x_i^2\;-\;\sum_i\;P_i\;x_i)^2}\)
  • Probability of getting 1 head in a single throw=\(\large\frac{1}{2}\);tail=\(\large\frac{1}{2}\)
P(x=0)=P(no tail)=\(P(H\;H\;H)\)
=\(\Large\frac{1}{2}\;\times\;\frac{1}{2}\;\times\;\frac{1}{2}\;=\;\frac{1}{8}\)
P(X=1)=P(1tail)
\(P(T\;T\:H\:,T\;H\;H\;,H\;T\;T)\)
 
=\(\Large\frac{1}{8}\;+\;\frac{1}{8}\;+\;\frac{1}{8}\;=\;\frac{3}{8}\)
P(X=2)=P(2tail)
\(P(H\;T\;T\;,T\;H\;T\;,T\;T\;H)\)
\(\Large\frac{1}{8}\;+\;\frac{1}{8}\;+\;\frac{1}{8}\;=\;\frac{3}{8}\)
 
P(X=3)=P(3tail)
\(P(T\;T\;T)\)
=\(\Large\;\frac{1}{8}\)
\(\Large\;\sum\;P_iX-i\;=\;\frac{1}{8}\;\times0+\;\frac{3}{8}\;\times1+\;\frac{3}{8}\;\times2+\;\frac{1}{8}\;\times3\)
\(\Large\;\frac{12}{8}\)
\(\Large\;\sum\;P_i\;X_i^2\;=\;\frac{1}{8}\;\times0+\;\frac{3}{8}\;\times1+\;\frac{3}{8}\;\times4+\;\frac{1}{8}\;\times9\)
\(\Large\frac{42}{8}\)
\(\Large\;\sigma=\sqrt{\frac{42}{8}-(\frac{12}{8})^2}\)=\(\Large\sqrt\frac{192}{64}\)
\(\Large\frac{\sqrt3}{2}\)

 

answered Feb 26, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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