Probability

In a pack of playing cards, all black kings and black queens are removed. If a card is drawn at random, find the probability of getting

1. a face card;


2. a black card;


3. a red card or a face card;


4. a black card or a face card

There are $2$ black kings and $2$ black queens which are removed from a pack of $52$ cards.

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 52 - 4 = 48$

1. Let $A =$ event of selecting a face card
   
   There are a total of $8$ face cards from which $4$ face cards are removed.
   
   $\therefore \;$ Number of elements in $A = n \left(A\right) = 4$
   
   $\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{4}{48} = \dfrac{1}{12}$


2. Let $B =$ event of selecting a black card
   
   There are $26$ black cards from which $4$ black cards (face cards) are removed.
   
   $\therefore \;$ Number of elements in $B = n \left(B\right) = 22$
   
   $\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{22}{48} = \dfrac{11}{24}$


3. Let $C = $ event of selecting a red card
   
   There are $26$ red cards.
   
   $\therefore \;$ Probability of drawing a red card $= P \left(\text{red card}\right) = \dfrac{26}{48}$
   
   $A =$ event of selecting a face card
   
   $P \left(A\right) = \dfrac{4}{48}$
   
   Let $X = $ event of selecting a red card OR a face card
   
   Probability of event $X = P \left(\text{C}\right) + P \left(\text{A}\right) = \dfrac{26}{48} + \dfrac{4}{48} = \dfrac{30}{48} = \dfrac{5}{8}$


4. Let $D = $ event of selecting a black card OR a face card
   
   Probability of event $D = P \left(\text{black card}\right) + P \left(\text{face card}\right) = \dfrac{22}{48} + \dfrac{4}{48} = \dfrac{26}{48} = \dfrac{13}{24}$