Question

Two cards are drawn successively without replacement from a well-shuffled pack of $52$ cards. The probability of drawing two aces is.
A: $\dfrac{1}{{26}}$
B: $\dfrac{1}{{221}}$
C: $\dfrac{4}{{223}}$
D: $\dfrac{1}{{13}}$

Answer 1

They have given them a pack of $52$ cards. In this we will be having $4$aces and $48$ others, but they are asking to find the probability of drawing two aces. By using the probability formula given by $P(A) = \dfrac{{n(A)}}{n}$ we can find the required answer.

Complete Step by Step Solution:
We have a well shuffled pack of $52$ cards, out of which we will be having $4$aces and $48$ other cards.
In order to find the probability of drawing two aces, we need to know the formula to find probability. So, the probability can be find by using the formula:
$P(A) = \dfrac{{n(A)}}{n}$
Where $P(A)$ denotes the probability of A
$n(A)$ is the number of occurrences of A or the number of favorable outcomes.
$n$ is the total number of possible outcomes or the sample space.

Let us consider $A$ be the number of aces drawn out of $52$ cards, that is $A = 2$
Now, to find probability of drawing two aces:
We have total four aces out of $52$ cards, which can be written as $P(A) = \dfrac{4}{{52}}$, when we draw first card then the probability becomes $P(A) = \dfrac{3}{{51}}$. By multiplying these two probabilities we get the required answer.
Therefore, $P(2) = \dfrac{4}{{52}} \times \dfrac{3}{{51}}$
Upon multiplication, we get
$\Rightarrow P(2) = \dfrac{{12}}{{2652}}$
Simplify or reduce the above equation to get the answer
$\Rightarrow P(2) = \dfrac{1}{{221}}$

Therefore, the probability of drawing two aces are $\dfrac{1}{{221}}$ and hence the option B is correct.

Note:
Here in this question they have asked for two aces, so we directly calculated for the same. They may ask to find the probability distribution of the number of aces, then you have to find the probability when $A = 0,1,2$. In probability questions one thing you need to learn is observation and analysation. When taking the sample space and the number of favorable outcomes we should take care that we are taking the correct value otherwise it will lead to the wrong answer.