A card is drawn at random from a deck of 52 cards. What is the probability that the drawn card is
- a queen or a club card
- a queen or a black card
$1$ card from $52$ cards can be selected in $52$ ways
$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 52$
Let,
$A$ be the event of drawing a queen card $\;\;\;$ (from $4$ queen cards);
$B$ be the event of drawing a club card $\;\;\;\;\;\;$ (from $13$ club cards);
$C$ be the event of drawing a black card $\;\;\;\;\;$ (from $26$ black cards).
Then, $n\left(A\right) = 4$, $n \left(B\right) = 13$, $n \left(C\right) = 26$
Also, $n \left(A \cap B\right) = 1$, $n \left(A \cap C\right) = 2$
- $P \left(\text{queen OR club}\right) = P \left(A \text{ OR } B\right) = P \left(A \cup B\right)$
$\because \;$ $A$ and $B$ are not mutually exclusive events, $\left(A \cap B\right) \neq 0$
$\begin{aligned} P \left(A \cup B\right) & = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} + \dfrac{n \left(B\right)}{n \left(S\right)} - \dfrac{n \left(A \cap B\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} \\\\ & = \dfrac{4}{13} \end{aligned}$
- $P \left(\text{queen OR a black card}\right) = P \left(A \text{ OR } C\right) = P \left(A \cup C\right)$
$\because \;$ $A$ and $C$ are not mutually exclusive events, $\left(A \cap C\right) \neq 0$
$\begin{aligned} P \left(A \cup C\right) & = P \left(A\right) + P \left(C\right) - P \left(A \cap C\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} + \dfrac{n \left(C\right)}{n \left(S\right)} - \dfrac{n \left(A \cap C\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} \\\\ & = \dfrac{7}{13} \end{aligned}$