Question

Two cards are drawn successively with replacement from a pack of 52 cards. Find the probability distribution of the number of aces. Find its mean and standard deviation.

Answer 1

Let us consider X as the number of aces. Since we can select 2 cards, both of them can be aces or 1 can be an ace or 0 ace. So the value of X can be 0, 1, 2. We have to find the probability of ace using the formula $P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ . Number of aces in a deck of cards will be 4. We also have to find the probability that the card drawn is not ace by subtracting the probability of ace from 1. Then, we have to find the probability at $X=0,1,2$ . $P\left( X=0 \right)$ implies the selection of two non-aces which will be the product of probability of non-ace twice. $X=1$ provides two cases: first one is ace while the second one is a non-ace and the first one as non-ace while the second one as ace. $P\left( X=1 \right)$ will be the sum of these. $X=2$ implies that the 2 cards are ace and $P\left( X=2 \right)$ can be found by multiplying the probability of ace twice. To find mean and standard deviation, we have to use the formula $E\left( x \right)=\mu =\sum\limits_{i=0}^{n}{{{x}_{i}}P\left( {{x}_{i}} \right)}$ and $\sigma =\sqrt{\sum\limits_{i=0}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}P\left( {{x}_{i}} \right)}}$ respectively.

Complete step-by-step solution:
We have to find the probability distribution of the number of aces, its mean and standard deviation. We are given a pack of 52 cards of which two cards are drawn successively with replacement.
Let X be the number of aces. We can select two cards. Of these, both of them can be aces or 1 can be an ace or 0 ace. So the value of X can be 0, 1, 2.
Out of 52 cards, we know that there are 4 aces. Therefore, we can find the probability of ace using the formula
$P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ .
$\Rightarrow P\left( \text{aces} \right)=\dfrac{4}{52}$
Now, let us find the probability that the card drawn is not a ace. For this, we have to subtract p(aces) from 1 since the probability of failure is $1-P\left( \text{success} \right)$ .
$\begin{aligned} & \Rightarrow P\left( \text{not ace} \right)=1-P\left( \text{ace} \right) \\\ & \Rightarrow P\left( \text{not ace} \right)=1-\dfrac{4}{52} \\\ & \Rightarrow P\left( \text{not ace} \right)=\dfrac{48}{52} \\\ \end{aligned}$
Now, let us consider the case when $X=0$ , that is, out of the two cards drawn, no ace is obtained. We can find the probability at $X=0$ by multiplying the probability of no ace twice since two cards are drawn.
$\begin{aligned} & \Rightarrow P\left( X=0 \right)=P\left( \text{not ace} \right)\times P\left( \text{not ace} \right) \\\ & \Rightarrow P\left( X=0 \right)=\dfrac{48}{52}\times \dfrac{48}{52} \\\ & \Rightarrow P\left( X=0 \right)=\dfrac{144}{169} \\\ \end{aligned}$
We have to consider the case of $X=1$ , that is, one card will be an ace and the other will be a non-ace. There can be two cases for this.
(i) The first one is ace while the second one is a non-ace
(ii) The first one as non-ace while the second one as ace.
We can find the probability at $X=1$ by adding the probabilities of (i) and (ii).
$\begin{aligned} & \Rightarrow P\left( X=1 \right)=P\left( \text{ ace} \right)\times P\left( \text{not ace} \right)+P\left( \text{not ace} \right)\times P\left( \text{ ace} \right) \\\ & \Rightarrow P\left( X=1 \right)=\dfrac{4}{52}\times \dfrac{48}{52}+\dfrac{48}{52}\times \dfrac{4}{52} \\\ & \Rightarrow P\left( X=1 \right)=\dfrac{24}{169} \\\ \end{aligned}$
Now, we have to consider the case of $X=2$ , that is, both the cards are aces.
$\begin{aligned} & \Rightarrow P\left( X=2 \right)=P\left( \text{ ace} \right)\times P\left( \text{ ace} \right) \\\ & \Rightarrow P\left( X=2 \right)=\dfrac{4}{52}\times \dfrac{4}{52} \\\ & \Rightarrow P\left( X=2 \right)=\dfrac{1}{169} \\\ \end{aligned}$
Therefore, we can tabulate the probability distribution of the number of aces as follows.

X	0	1	2
P(X)	$\dfrac{144}{169}$	$\dfrac{24}{169}$	$\dfrac{1}{169}$

Now, let us find the mean and standard deviation of the probability distribution of the number of aces. We know that mean is given by
$E\left( x \right)=\mu =\sum\limits_{i=0}^{n}{{{x}_{i}}P\left( {{x}_{i}} \right)}$
Here, $n=3$ . Therefore, we can find the required mean as follows.
$\begin{aligned} & \Rightarrow \mu =0\times \dfrac{144}{169}+1\times \dfrac{24}{169}+2\times \dfrac{1}{169} \\\ & \Rightarrow \mu =0+0.142+0.118 \\\ & \Rightarrow \mu =0.26 \\\ \end{aligned}$
Now, let us find the standard deviation using the formula
$\sigma =\sqrt{\sum\limits_{i=0}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}P\left( {{x}_{i}} \right)}}$
$\begin{aligned} & \Rightarrow \sigma =\sqrt{{{\left( 0-0 \right)}^{2}}\times \dfrac{144}{169}+{{\left( 1-0.26 \right)}^{2}}\times \dfrac{24}{169}+{{\left( 2-0.26 \right)}^{2}}\times \dfrac{1}{169}} \\\ & \Rightarrow \sigma =\sqrt{0.077+0.179} \\\ & \Rightarrow \sigma =\sqrt{0.256} \\\ & \Rightarrow \sigma =0.505 \\\ \end{aligned}$
Hence the mean and standard deviation is 0.26 and 0.505 respectively.

Note: Students must know the number of aces, kings, queens, jack, diamond, heart and spheres in a deck of cards. They must carefully find the probability at $X=0,1,2$ since there can be a chance of making a mistake by adding the probabilities instead of multiplying. Students must be thorough with the formulas of mean, variance and standard deviation. They must note that standard deviation is the square root of variance.