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)

{{ = ^{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.