# 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.

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

edited Jun 4, 2013