Question

From a pack of $52$ cards, two cards are drawn in succession one by one without replacement. The probability that both are aces is
A) $\dfrac{2}{13}$
B) $\dfrac{1}{51}$
C) $\dfrac{1}{221}$
D) $\dfrac{2}{21}$

Answer 1

Hint: In this question we have to find the probability of getting two aces in succession from a pack of $52$ cards. There are a total of $4$ aces in a well shuffled pack of $52$ cards in which $2$ are black and $2$ are red. We will use this data and basic formula of probability and combination to get our required answer.

Complete step-by-step answer:
First we will learn some concepts of probability to understand the question easily.
Random Experiment: It is an experiment in which we know all the possible outcomes in advance but we cannot predict any specific outcome before the completion of the task.
Example: Tossing a coin, rolling a dice….
Sample Space: It is a set of all possible outcomes of a random experiment.
If we toss a coin we can get head (H) or tail (T)
So the sample space of tossing a coin is: {H, T}
Event: Event is a subset of a sample space of a random experiment
Sample space of tossing a coin is (H, T}. Getting a head (H) or tail (T) is an event.
Let us learn these definitions using examples of rolling a dice also.
Rolling a dice is a random experiment. We can get $1,2,3,4,5,6$ as an outcome so sample space is
\[\{1,2,3,4,5,6\}\] .
$\{1\}$ Is a subset of \[\{1,2,3,4,5,6\}\] so getting $1$ by rolling a dice is an event
Probability is a chance of getting our required event or getting our favourable outcomes
 Probability of getting an event E is
$\Rightarrow P\left( E \right)=\dfrac{No.\text{ of favourable outcomes}}{Total\text{ no}\text{. of outcomes}}$
Now we will learn about a pack of $52$ cards
There are a total of $4$ suits in a pack of $52$ cards and each suit has $13$ cards.
Out of $4$ suits, $2$ are black in colour and $2$ are red in colour.
$4$ Suits are named as Heart, Diamond, Club, and Spades
Heart and Diamond are red in colour & Club and Spade are black in colour.
In each suite there are $4$ Kings, $4$ Queens, $4$ Aces and $4$ Jacks.
In which $2$ Kings, $2$ Queens, 2 Aces and $2$ Jacks are red in colour and $2$ Kings, $2$ Queens, 2 Aces and $2$ Jacks are in black colour.
So in a pack of $52$ cards there are \[4\]Aces.
So no. of possibilities of getting \[2\] Aces without replacement $=$ \[^{4}{{C}_{2}}\]
Total no. of possibilities of selecting \[2\] cards without replacement $={}^{52}{{C}_{2}}$
As we know, probability of event E,
  $\Rightarrow P\left( E \right)=\dfrac{\text{No}\text{. of favourable outcomes}}{\text{Total no}\text{. of possibilities }}$
Probability of getting $2$ aces without replacement is given as:
$\Rightarrow P(E)=\dfrac{^{4}{{C}_{2}}}{^{52}{{C}_{2}}}$
Using the formula,
${{\Rightarrow }^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Putting the values in the above formula:
$\begin{align}
  & \Rightarrow \dfrac{\dfrac{4!}{2!\left( 4-2 \right)!}}{\dfrac{52!}{2!\left( 52-2 \right)!}} \\
 & \Rightarrow \dfrac{\dfrac{4!}{2!\times 2!}}{\dfrac{52!}{2!\times 50!}} \\
\end{align}$
Probability of getting $2$ aces without replacement
Using the formula given below
$\Rightarrow n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).......\left( n-r \right)!$ where $n\ge r$
$\begin{align}
  & \Rightarrow \dfrac{\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}{\dfrac{52\times 51\times 50!}{2\times 1\times 50!}} \\
 & \Rightarrow \dfrac{\dfrac{6}{1}}{\dfrac{52\times 51}{2}} \\
 & \Rightarrow \dfrac{6}{26\times 51} \\
 & \\
\end{align}$
Probability of getting $2$ Aces without replacement
$\begin{align}
  & \Rightarrow \dfrac{1}{13\times 17} \\
 & \Rightarrow \dfrac{1}{221} \\
 & \\
\end{align}$
Therefore, the correct answer is Option (C).

Note: Range of Probability of any event is $\left[ 0,1 \right]$ . If there are two events A and B such that they cannot occur at same time or we can say that both events are independent then we will use product rule $P\left( A.B \right)=P\left( A \right).P\left( B \right)$ and when events A and B can occur at same time then we will use addition rule $P\left( A+B \right)=P\left( A \right)+P\left( B \right)$ .