Question

Four cards are drawn from a pack of 52 cards what is the chance that cards are king queen jack and ace

Answer 1

$\underline{\Huge{\textbf{About\: A\: Pack\: of\: Cards}}}$

(Refer Attachment)
In a Pack of 52 Cards,
There are 4 Suits viz., Clubs, Spades, Hearts, Diamonds
Each of Suit consists 13 cards viz., A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.

So,
$\underline{\textsf{There are 4 Kings, 4 Queens, 4 Jacks and 4 Aces in a pack of cards}}$

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
Experiment - 4 cards are drawn from a Pack of 52 Cards

Probability that drawn cards are King, Queen, Jack and Ace = ?

$\underline{\Large{\textbf{Assumptions\: and Cases:}}}$

Assuming that the Experiment is done $\underline{without Replacement}$
(Without Replacement means the card drawn is not replaced (put back) in the pack of 52 cards)
[ It is necessary to $\textsf{track the total}$ when going through Experiments without Replacements]

Here,
There are $\large{\textsf{Two cases}}$ for the Happening of Event(Drawn 4 cards are King, Queen, Jack, Ace) depending on the order of drawing.

$\textbf{Case-(1):}$ If there is Criterion on Event about $\textit{Sequence}$

$\textbf{Case-(2):}$ If there is $\textit{No}$ Criterion on Event about $\textit{Sequence}$

Let S be the Sample space

Let $E_{1}$ be the Event that First card drawn being King.

Let $E_{2}$ be the Event that Second card drawn being Queen.

Let $E_{3}$ be the Event that First card drawn being Jack.

Let $E_{4}$ be the Event that First card drawn being Ace.

Let $P_{1}$ be the Probability that first card being king , second card being Queen, Third card being Jack, Fourth card being Ace

Let $P_{2}$ be the Probability that the drawn 4 cards consists King, Queen, Jack and Ace

We have,
$\bigstar \textsf{Probability of Event = \dfrac{No. \: of \: favorable \: outcomes}{Total \: No. \: of \: outcomes} }$

$\underline{\Large{\textbf{Solution\: for\: case-1:}}}$

$\underline{\textit{Find Probability of Event E_{1} }}$

$P(E_{1}) = \dfrac{n(E_{1})}{n(S)}$

$P(E_{1}) = \dfrac{4}{52}$

(Total cards left = 51)

$\underline{\textit{Find Probability of Event E_{2} }}$

$P(E_{2}) = \dfrac{n(E_{2})}{n(S)}$

$P(E_{2}) = \dfrac{4}{51}$

(Total cards left = 50)

$\underline{\textit{Find Probability of Event E_{3} }}$

$P(E_{3}) = \dfrac{n(E_{3})}{n(S)}$

$P(E_{3}) = \dfrac{4}{50}$

(Total cards left = 49)

$\underline{\textit{Find Probability of Event E_{4} }}$

$P(E_{4}) = \dfrac{n(E_{4})}{n(S)}$

$P(E_{4}) = \dfrac{4}{49}$

$\underline{\textit{Find Probability P_{1} }}$

$P_{1}= P(E_{1}) \times P(E_{2}) \times P(E_{3}) \times P(E_{4})$

$\implies P_{1} = \dfrac{4}{52} \times \dfrac{4}{51} \times \dfrac{4}{50} \times \dfrac{4}{49}$

$\implies P_{1} = \dfrac{32}{81275}$

$\underline{\Large{\textbf{Solution\: for\: case-2:}}}$

If there is no criterion on the order of drawing,
the 4 cards King, Queen, Jack and Ace can be $\textsf{drawn in any order}$.

$\textit{Find No. of ways to Order 4 cards}$

We have,
$\bigstar \textsf{No. of ways of arranging 'n' distinct objects = n!}$

$\implies$ No. of ways to Order 4 cards = $\textbf{4! = 24}$

$\textit{Find Probability P_{2} }$

$P_{2} = (P_{1}) \times (No.\: of\: ways\: of\: Arranging\: 4\: cards)$

$P_{2} = \dfrac{32}{81275} \times 24$

$\implies P_{2} = \dfrac{768}{81275}$

$\therefore$

$\textsf{Probability that first card being king, second card being Queen, Third card being Jack, Fourth card being Ace = \underline{ \dfrac{32}{81275} }}$

$\textsf{Probability that the drawn 4 cards consists King, Queen, Jack and Ace = \underline{ \dfrac{768}{81275} }}$