Question

Two cards are drawn simultaneously from a pack of \[52\] cards. Compute the mean and standard deviation (S.D) of the number of kings.

Answer 1

Hint: To solve this question, first we will start with the probability of getting a number of kings. Then using the values we will make a probability distribution table and then we will find mean and standard deviation (S.D) of the number of kings.

Complete step-by-step answer:
We have been given a pack of \[52\] cards. And we know that a pack has \[4\] kings.
Now, let X be the number of kings we obtained.
It is given that two cards are drawn simultaneously, so we can get \[0,1\] or \[2\] kings.
This means the value of X can be \[0,1\] or \[2.\]
So, to get the total number of ways to draw \[2\] cards out of 52, we will use the formula mentioned below.
\[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
where, n = number of items
r = number of items being chosen at a time
Now, total number of ways to draw \[2\] cards out of \[52{\text{ }}{ = ^{52}}{C_2}\]
\[ = {\text{ }}1326\]
Now let us check the probability of getting \[0,1\] or \[2\] kings.
Probability of getting 0 kings \[ = {\text{ }}P\left( {X = 0} \right)\]
Number of ways of getting \[0\] kings \[ = \] Number of ways of drawing 2 cards out of non-king cards (\[52 - 4,\]i.e., \[48\]cards)
\[\begin{array}{*{20}{l}}
  {{ = ^{48}}{C_2}} \\
  { = {\text{ }}1128}
\end{array}\]
So, \[P\left( {X = 0} \right){\text{ }} = \]$\dfrac{{Number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ }}0{\text{ }}kings}}{{Total{\text{ }}number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ }}kings}}$
$ = \dfrac{{1128}}{{1326}}$
Probability of getting \[1\] king \[ = {\text{ }}P\left( {X = 1} \right)\]
Number of ways of getting \[1\] king \[ = \] Number of ways of drawing \[1\] king out of \[4\] king cards \[ \times \]Number of ways of drawing \[1\] card out of \[48\] non-king \[48\] cards
\[\begin{array}{*{20}{l}}
  {{ = ^4}{C_1}{ \times ^{48}}{C_1}} \\
  { = {\text{ }}4 \times 48} \\
  { = {\text{ }}192}
\end{array}\]
So, \[P\left( {X = 1} \right){\text{ }} = \]$\dfrac{{Number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ 1 }}king}}{{Total{\text{ }}number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ }}kings}}$
$ = \dfrac{{192}}{{1326}}$
Now, Probability of getting \[2\] kings \[ = {\text{ }}P\left( {X = 2} \right)\]
Number of ways of getting \[2\] kings \[ = \] Number of ways of drawing \[2\] kings out of \[4\] king cards
\[\begin{array}{*{20}{l}}
  {{ = ^4}{C_2}} \\
  { = {\text{ }}6}
\end{array}\]
So, \[P\left( {X = 2} \right){\text{ }} = \]$\dfrac{{Number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ 2 }}kings}}{{Total{\text{ }}number{\text{ }}of{\text{ }}ways{\text{ }}of{\text{ }}getting{\text{ }}kings}}$
$ = \dfrac{6}{{1326}}$
So, we get the probability of \[0,{\text{ }}1\] and \[2\] kings. Now we will make the probability distribution of it. So, the probability distribution table is,

X	\[0\]	\[1\]	\[2\]
P(X)	$\dfrac{{1128}}{{1326}}$	$\dfrac{{192}}{{1326}}$	$\dfrac{6}{{1326}}$

Now, we need to find the mean.
We know that, $Mean\overline X = \sum {XP(X)} $
\[
   = 0 \times \dfrac{{1128}}{{1326}} + 1 \times \dfrac{{192}}{{1326}} + 2 \times \dfrac{6}{{1326}} \\
   = \dfrac{{192}}{{1326}} + \dfrac{{12}}{{1326}} \\
   = \dfrac{{204}}{{1326}} \\
   = \dfrac{{34}}{{221}} \\
\]
\[\begin{array}{*{20}{l}}
  {\therefore Mean{\text{ }} = {\text{ }}\dfrac{{34}}{{221}}} \\
  {}
\end{array}\]
Now, we need to find the SD for which we need to find variance first.
We know that, \[Variance{\text{ }} = \sum {X^2}{\text{P(X) - }}\overline {{{\text{X}}^2}} \]
\[
   = [{0^2} \times \dfrac{{1128}}{{1326}} + {1^2} \times \dfrac{{192}}{{1326}} + {2^2} \times \dfrac{6}{{1326}}] - {(\dfrac{{34}}{{221}})^2} \\
   = [\dfrac{{192}}{{1326}} + \dfrac{{24}}{{1326}}] - {(\dfrac{{34}}{{221}})^2} \\
   = \dfrac{{216}}{{1326}} - {(\dfrac{{34}}{{221}})^2} \\
   = \dfrac{{36}}{{221}} - {(\dfrac{{34}}{{221}})^2} \\
   = \dfrac{1}{{221}}(36 - \dfrac{{{{34}^2}}}{{221}}) \\
   = \dfrac{1}{{221}}[\dfrac{{221 \times 36 - 1156}}{{221}}] \\
   = \dfrac{1}{{221}}[\dfrac{{6800}}{{221}}] \\
   = \dfrac{{6800}}{{{{(221)}^2}}} \\
    \\
\]
\[\begin{array}{*{20}{l}}
  {\therefore Variance{\text{ }} = {\text{ }}\dfrac{{6800}}{{{{221}^2}}}} \\
  {}
\end{array}\]
We know that, \[SD = {\sigma _x} = \sqrt {Variance} \]
So, \[\begin{array}{*{20}{l}}
  {SD{\text{ }} = {\text{ }}\sqrt {\dfrac{{6800}}{{{{221}^2}}}} } \\
  {}
\end{array}\]
$
   = \dfrac{{\sqrt {6800} }}{{221}} \\
   = \dfrac{{82.46}}{{221}} \\
  \therefore SD = 0.37 \\
$
Thus, the mean and standard deviation of the number of kings is $\dfrac{{34}}{{221}}$ and \[0.37\] respectively.
Note: The formula which we have used above in the solutions i.e., \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] . This formula is used to calculate combinations. In this formula, n represents total number of items and r represents the number of items being chosen.