logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

Two cards are drawn successively without replacement from a well shuffled deck of cards.Find the mean and standard variation of the random variable X where X is the number of aces.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \(A\;pack\;of\;52\;cards\;contain\;4\;aces\)
  • \(In\;a;toss\;of\;dies\;twice\)
  • \(X=\;no\;of\;aces\)
  • \(X=(0\;1\;2\;)\)
  • \(Mean\;=\sum\;P_i\;X_i\)
  • \(\sigma\;=\sqrt{variance}=\sqrt{\sum\;P_iX^2_i\;-\;(\sum\;P_iX_i)^2}\)
\(P(X=0)=p(no\;aces)\)
\(\Large\frac{48}{52}\times\frac{47}{51}\)
\(P(X=1)=p(1\;aces/no\;aces)\)
\(\Large\;2(\frac{48}{52}\times\frac{47}{51})\)
\(P(X=2)=p(both\;aces)\)
\(\Large\frac{4}{52}\times\frac{3}{51}\)
\(E(X)=\sum\;P_iX_i\)
=\(\Large\;0\times\frac{48\times47}{52\times51}\;+\;1\times2\times\frac{4}{52}\times\frac{48}{51}+2\times\frac{4\times3}{52\times51}\)
=\(\Large\;\frac{204}{26\times51}\;=\frac{2}{13}\)
\(\sum\;P_iX_i^2=\Large\;1^2\;\times\;2\;\times\;\frac{4}{52}\times\frac{48}{51}+2^2\times\frac{4\times3}{52\times51}\)
\(var\;=\;0.2952-(.153)^2\)
=\(0.1421\)
\(\sigma\;=\sqrt{0.1421}=0.377\)

 

answered Mar 17, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...