Question

Two Cards are drawn without replacement from a well-shuffled pack. Find the probability that one of them is an ace of heart.
A. $\dfrac{1}{{25}}$
B. $\dfrac{1}{{26}}$
C. $\dfrac{1}{{52}}$
D. None of these

Answer 1

Hint: In the case of probability without replacement, the events occurring are referred to as dependent events i.e., in the given problem, the second event is dependent on the first event. Therefore, we need to evaluate the probability of the first event only after which the required probability will be calculated as per the question.

Formula Used:
The basic formula used for evaluating probability here is:
$Probability{\text{ }}of{\text{ }}getting{\text{ }}an{\text{ }}outcome = \dfrac{{Number{\text{ }}of{\text{ }}getting{\text{ }}favorable{\text{ }}outcome}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}outcomes}}$

Complete step by step solution: 
We know that, the total number of cards in a well-shuffled pack $(N) = 52$cards.
And, in a pack of 52 cards, there is only $1$ Ace of Heart and the remaining 51 are non-Ace of Heart.
Let ${E_1}$be the event of getting an Ace of Heart and ${E_2}$ be the event of getting non-Ace of Heart cards.
Now, we know $P = \dfrac{E}{N}$, where,
P = Probability of getting an outcome
E = Number of getting favorable outcomes
N = Total Number of outcomes
Therefore, the probability of getting an Ace of Heart is $P({E_1}) = \dfrac{{{E_1}}}{N} = \dfrac{1}{{52}}$
And, Probability of getting non-Ace of Heart is $P({E_2}) = \dfrac{{{E_2}}}{N} = \dfrac{{51}}{{52}}$

Now, According to Question (ATQ), there will be 2 cases (illustrated in the figure): -
Case1. When 1^st card is drawn is ACE of HEART and the 2nd will be non-ACE of HEART
Probability (of getting the first card is ace of heart and the second is non-ace of heart) = $P(A) = \dfrac{1}{{52}} \times \dfrac{{51}}{{51}} = \dfrac{1}{{52}}$
Case2. When 1^st card is drawn is non-ACE of HEART and the 2nd will be ACE of HEART
Probability (of getting the first card is non-ace of heart and the second is ace of heart) = $P(B) = \dfrac{{51}}{{52}} \times \dfrac{1}{{51}} = \dfrac{1}{{52}}$
Thus, the required probability is:
$P(E) = P(A) + P(B) = \dfrac{1}{{52}} + \dfrac{1}{{52}} = \dfrac{2}{{52}} = \dfrac{1}{{26}}$
Hence, the correct option is: (B) $\dfrac{1}{{26}}$.

Note: Since the problem is based on Probability without replacement which means an item cannot be drawn more than once hence, it is very necessary to analyze the given conditions carefully and the study of both the cases must be done precisely to find the accurate solution. Calculations must be performed very carefully.