Question

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of kings.

Answer 1

Hint: For solving this question, we use the concept of Binomial Distribution. Using the formula $P\left( X=x \right)={}^{n}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}$, we can easily the calculate the probability distribution of the number of kings drawn from the deck of cards.
Complete step-by-step answer:
The Binomial Distribution is used when there are exactly two mutually exclusive outcomes of trials. The outcomes of the event are labelled as success and failure. And the formula of Binomial Distribution is:
$P\left( X=x \right)={}^{n}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}$
Here, x represents success and n represent number of trials, p is the probability of success on a single trial and q is the probability of failure on a single trial.
Number of cards drawn = n = 2.
Probability of a getting a king = p $=\dfrac{4}{52}=\dfrac{1}{13}$
Here, p is success.
Now, p + q = 1, so the probability of failure could be expressed as:
q = 1 – p $=1-\dfrac{1}{13}=\dfrac{12}{13}$
$\Rightarrow P\left( X=x \right)={}^{2}{{C}_{x}}{{\left( \dfrac{12}{13} \right)}^{n-x}}{{\left( \dfrac{1}{13} \right)}^{x}}....(1)$
Two cards are drawn from the deck of 52 cards,
We can get 0 king, 1 king or 2 kings.
So, the values of X can be 0,1 and 2.
Putting the value of x = 0 in equation (1), we get
$\begin{aligned} & P\left( X=0 \right)={}^{2}{{C}_{0}}{{\left( \dfrac{12}{13} \right)}^{2-0}}{{\left( \dfrac{1}{13} \right)}^{0}} \\\ & P\left( X=0 \right)=1\times 1\times {{\left( \dfrac{12}{13} \right)}^{2}}=\dfrac{144}{169} \\\ \end{aligned}$
Hence, the probability of getting no king is $\dfrac{144}{169}$.
$\begin{aligned} & P\left( X=1 \right)={}^{2}{{C}_{1}}{{\left( \dfrac{12}{13} \right)}^{2-1}}{{\left( \dfrac{1}{13} \right)}^{1}} \\\ & P\left( X=1 \right)=2\times {{\left( \dfrac{1}{13} \right)}^{1}}\times {{\left( \dfrac{12}{13} \right)}^{1}} \\\ & P\left( X=1 \right)=\dfrac{24}{169} \\\ \end{aligned}$
Hence, the probability of getting one king is $\dfrac{24}{169}$.
$\begin{aligned} & P\left( X=2 \right)={}^{2}{{C}_{2}}{{\left( \dfrac{12}{13} \right)}^{2-2}}{{\left( \dfrac{1}{13} \right)}^{2}} \\\ & P\left( X=2 \right)=1\times {{\left( \dfrac{1}{13} \right)}^{2}} \\\ & P\left( X=2 \right)=\dfrac{1}{169} \\\ \end{aligned}$
Hence, the probability of getting two kings are $\dfrac{1}{169}$.
Note: The key concept involved in solving this problem is the knowledge of a binomial distribution. Students can also verify their result by summing all the probabilities. For the above-mentioned case, when we add all the probability, we get 1. Hence, our obtained result is correct.