Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards?
(a) Mean = 0.1 and Variance = 0.7
(b) Mean = 0.6 and Variance = 0.3
(c) Mean = 0.49 and Variance = 0.37
(d) Mean = 1 and variance = 0.49
Answer
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Hint: We start solving the problem by recalling the total number of black and red cards present in a pack of cards. We then find the probability to draw 2 black, 1 black and 1 red, 2 red cards without replacement from the given pack of 52 cards. We then use mean = $\sum\limits_{r=0}^{n}{rP\left( X=r \right)}$ and variance = $\left( \sum\limits_{r=0}^{n}{{{r}^{2}}P\left( x=r \right)} \right)-{{\left( \text{Mean} \right)}^{2}}$ and make the necessary calculations to get the required result.
Complete step-by-step solution:
According to the problem, it is given that we have drawn two cards without replacement from a well-shuffled pack of 52 cards. We need to find the mean and variance of red cards.
We know that there will 26 red cards and 26 black cards in a pack of 52 cards.
Let us find the probability that we draw 2 black cards.
So, we get $P\left( B=2 \right)=\dfrac{\text{no}\text{. of ways to draw 2 cards out of 26 black cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( B=2 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{26!}{2!\left( 26-2 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{26!}{2!24!}}{\dfrac{52!}{2!50!}}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{26\times 25}{52\times 51}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{25}{2\times 51}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{25}{102}$ ---(1).
Let us find the probability that we draw 1 black card and 1 red card.
So, we get $P\left( B=1,R=1 \right)=\dfrac{\text{no}\text{. of ways to draw 1 cards out of 26 black cards}\times \text{no}\text{. of ways to draw 1 cards out of 26 red cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{{}^{26}{{C}_{1}}\times {}^{26}{{C}_{1}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{\dfrac{26!}{1!\left( 26-1 \right)!}\times \dfrac{26!}{1!\left( 26-1 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{\dfrac{26!}{1!\left( 25 \right)!}\times \dfrac{26!}{1!\left( 25 \right)!}}{\dfrac{52!}{2!\left( 50 \right)!}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{26\times 26}{\dfrac{52\times 51}{2\times 1}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{26}{51}$ ---(2).
Let us find the probability that we draw 2 red cards.
So, we get $P\left( R=2 \right)=\dfrac{\text{no}\text{. of ways to draw 2 cards out of 26 red cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( R=2 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( R=2 \right)=\dfrac{\dfrac{26!}{2!\left( 26-2 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{\dfrac{26!}{2!24!}}{\dfrac{52!}{2!50!}}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{26\times 25}{52\times 51}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{25}{2\times 51}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{25}{102}$ ---(3).
We know that the mean is defined as $\sum\limits_{r=0}^{n}{rP\left( X=r \right)}$ and variance is defined as $\left( \sum\limits_{r=0}^{n}{{{r}^{2}}P\left( x=r \right)} \right)-{{\left( \text{Mean} \right)}^{2}}$.
So, let us find the mean of the number of red cards according to the problem.
So, we get $\text{mean}=\sum\limits_{r=0}^{2}{rP\left( X=r \right)}$, where r is no. of red cards.
$\Rightarrow \text{mean}=\left( 0\times P\left( X=0 \right) \right)+\left( 1\times P\left( X=1 \right) \right)+\left( 2\times P\left( X=2 \right) \right)$.
From equations (1), (2), and (3) we get
$\Rightarrow \text{mean}=0+\left( 1\times \dfrac{26}{51} \right)+\left( 2\times \dfrac{25}{102} \right)$.
$\Rightarrow \text{mean}=\dfrac{26}{51}+\dfrac{25}{51}$.
$\Rightarrow \text{mean}=\dfrac{51}{51}$.
$\Rightarrow \text{mean}=1$.
Now, we have $\text{variance=}\left( \sum\limits_{r=0}^{2}{{{r}^{2}}P\left( x=r \right)} \right)-{{\left( \text{Mean} \right)}^{2}}$, where r is no. of red cards.
$\Rightarrow \text{variance=}\left( {{0}^{2}}\times P\left( x=0 \right) \right)+\left( {{1}^{2}}\times P\left( x=1 \right) \right)+\left( {{2}^{2}}\times P\left( x=2 \right) \right)-{{\left( \text{1} \right)}^{2}}$.
$\Rightarrow \text{variance=}0+\left( 1\times \dfrac{26}{51} \right)+\left( 4\times \dfrac{25}{102} \right)-\text{1}$.
$\Rightarrow \text{variance=}\dfrac{26}{51}+\dfrac{100}{102}-1$.
$\Rightarrow \text{variance=0}\text{.51}+0.98-1$.
$\Rightarrow \text{variance=0}\text{.49}$.
So, we have found mean and variance as 1 and 0.49.
The correct option for the given problem is (d).
Note: According to the problem, we can see that we need to find mean and variance for the total no. of red cards. This means that we have found the total possibilities of occurrence of red cards while drawing two cards from the pack. We should not confuse suits with the colors of cards in the pack. We can also find the standard deviation which is the square root of the variance. Similarly, we can also expect problems to find the mean and variance of the kings, queens while drawing cards.
Complete step-by-step solution:
According to the problem, it is given that we have drawn two cards without replacement from a well-shuffled pack of 52 cards. We need to find the mean and variance of red cards.
We know that there will 26 red cards and 26 black cards in a pack of 52 cards.
Let us find the probability that we draw 2 black cards.
So, we get $P\left( B=2 \right)=\dfrac{\text{no}\text{. of ways to draw 2 cards out of 26 black cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( B=2 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{26!}{2!\left( 26-2 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{\dfrac{26!}{2!24!}}{\dfrac{52!}{2!50!}}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{26\times 25}{52\times 51}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{25}{2\times 51}$.
$\Rightarrow P\left( B=2 \right)=\dfrac{25}{102}$ ---(1).
Let us find the probability that we draw 1 black card and 1 red card.
So, we get $P\left( B=1,R=1 \right)=\dfrac{\text{no}\text{. of ways to draw 1 cards out of 26 black cards}\times \text{no}\text{. of ways to draw 1 cards out of 26 red cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{{}^{26}{{C}_{1}}\times {}^{26}{{C}_{1}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{\dfrac{26!}{1!\left( 26-1 \right)!}\times \dfrac{26!}{1!\left( 26-1 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{\dfrac{26!}{1!\left( 25 \right)!}\times \dfrac{26!}{1!\left( 25 \right)!}}{\dfrac{52!}{2!\left( 50 \right)!}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{26\times 26}{\dfrac{52\times 51}{2\times 1}}$.
$\Rightarrow P\left( B=1,R=1 \right)=\dfrac{26}{51}$ ---(2).
Let us find the probability that we draw 2 red cards.
So, we get $P\left( R=2 \right)=\dfrac{\text{no}\text{. of ways to draw 2 cards out of 26 red cards}}{\text{no}\text{. of ways to draw 2 cards out of total 52 cards}}$.
We know that the total no. of ways to draw r items out of given n items without replacement is ${}^{n}{{C}_{r}}$ ways.
$\Rightarrow P\left( R=2 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times ......\times 3\times 2\times 1$.
$\Rightarrow P\left( R=2 \right)=\dfrac{\dfrac{26!}{2!\left( 26-2 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{\dfrac{26!}{2!24!}}{\dfrac{52!}{2!50!}}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{26\times 25}{52\times 51}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{25}{2\times 51}$.
$\Rightarrow P\left( R=2 \right)=\dfrac{25}{102}$ ---(3).
We know that the mean is defined as $\sum\limits_{r=0}^{n}{rP\left( X=r \right)}$ and variance is defined as $\left( \sum\limits_{r=0}^{n}{{{r}^{2}}P\left( x=r \right)} \right)-{{\left( \text{Mean} \right)}^{2}}$.
So, let us find the mean of the number of red cards according to the problem.
So, we get $\text{mean}=\sum\limits_{r=0}^{2}{rP\left( X=r \right)}$, where r is no. of red cards.
$\Rightarrow \text{mean}=\left( 0\times P\left( X=0 \right) \right)+\left( 1\times P\left( X=1 \right) \right)+\left( 2\times P\left( X=2 \right) \right)$.
From equations (1), (2), and (3) we get
$\Rightarrow \text{mean}=0+\left( 1\times \dfrac{26}{51} \right)+\left( 2\times \dfrac{25}{102} \right)$.
$\Rightarrow \text{mean}=\dfrac{26}{51}+\dfrac{25}{51}$.
$\Rightarrow \text{mean}=\dfrac{51}{51}$.
$\Rightarrow \text{mean}=1$.
Now, we have $\text{variance=}\left( \sum\limits_{r=0}^{2}{{{r}^{2}}P\left( x=r \right)} \right)-{{\left( \text{Mean} \right)}^{2}}$, where r is no. of red cards.
$\Rightarrow \text{variance=}\left( {{0}^{2}}\times P\left( x=0 \right) \right)+\left( {{1}^{2}}\times P\left( x=1 \right) \right)+\left( {{2}^{2}}\times P\left( x=2 \right) \right)-{{\left( \text{1} \right)}^{2}}$.
$\Rightarrow \text{variance=}0+\left( 1\times \dfrac{26}{51} \right)+\left( 4\times \dfrac{25}{102} \right)-\text{1}$.
$\Rightarrow \text{variance=}\dfrac{26}{51}+\dfrac{100}{102}-1$.
$\Rightarrow \text{variance=0}\text{.51}+0.98-1$.
$\Rightarrow \text{variance=0}\text{.49}$.
So, we have found mean and variance as 1 and 0.49.
The correct option for the given problem is (d).
Note: According to the problem, we can see that we need to find mean and variance for the total no. of red cards. This means that we have found the total possibilities of occurrence of red cards while drawing two cards from the pack. We should not confuse suits with the colors of cards in the pack. We can also find the standard deviation which is the square root of the variance. Similarly, we can also expect problems to find the mean and variance of the kings, queens while drawing cards.
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