Question

A card is drawn randomly from the deck of cards. Then, the probability of this card is red or a queen is
A. $\dfrac{1}{13}$
B. $\dfrac{1}{26}$
C. $\dfrac{1}{2}$
D. $\dfrac{7}{13}$

Answer 1

Hint: In this question, we are to find the probability of a card drawn randomly being a red or queen. Here, the addition theorem on probability is applied for finding the required probability.

Formula used: The probability is calculated by,
$P(E)=\dfrac{n(E)}{n(S)}$
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A+B)=P(A)+P(B)-P(A\cap B)$
When two events happen independently, the occurrence of one is not impacted by the occurrence of the other.
For the events $A$ and $B$, $P(A\cap B)=P(A)\cdot P(B)$ if they are independent and $P(A\cap B)=\Phi $ if they are mutually exclusive.

Complete step by step solution: A pack of cards contains \[52\] cards. $26$ are red cards and the other $26$ are black cards.
 There are $4$ queens, kings, Aces and Number cards (1 – 10). There are $13$ diamonds, clubs, hearts and diamonds.
It is given that, a card is drawn randomly.
Then, the probability of receiving a red card is
$P(R)=\dfrac{26}{52}$
The probability of getting a queen card in a draw:
$P(Q)=\dfrac{4}{52}$
The probability of receiving a red queen card:
$P(R\cap Q)=\dfrac{2}{52}$
Then, the required probability of drawing a card which is red or a queen is
$\begin{align}
  & P(A+B)=P(A)+P(B)-P(A\cap B) \\
 & \text{ }=\dfrac{26}{52}+\dfrac{4}{52}-\dfrac{2}{52} \\
 & \text{ =}\dfrac{7}{13} \\
\end{align}$

Thus, Option (D) is correct.

Note: In this question, the addition theorem on probability is applied for finding the required probability. By substituting the appropriate values, the required probability is calculated. Here we may go wrong with the count of cards. There are fifty-two cards half red coloured and the other half black colour.