Question

A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is,
(i) a black king (ii) either a black card or a king (iii) a jack, queen or a king (iv) neither an ace nor a king (v) spade or an ace (vi) neither a red card nor a queen (vii) other than an ace (viii) a ten (ix) a spade (x) a black card (xi) the seven of clubs (xii) jack (xiii) the ace of spades (xiv) a queen (xv) a heart (xvi) a red card (xvii) neither a king nor a queen.

Answer 1

Hint: There are 52 cards in a deck. The probability of finding a card can be calculated by dividing the number of cards of the given type by the total number of cards.

Complete step-by-step answer:
Total number of cards (T) = 52
(i) a black king:
In a deck of cards, there are two black kings, one of spade and one of clubs.
\[\therefore \] Probability of finding a black king \[=\dfrac{2}{25}=\dfrac{1}{13}\]
(ii) either a black card or a king:
Total black cards = 13 x 2 = 26
Total kings other than black cards = 2
\[\therefore \] Probability $=\dfrac{2\times 26}{52}=\dfrac{28}{52}=\dfrac{7}{13}$
(iii) a jack, a queen or a king:
Total jacks = 4
Total queens = 4
Total kings = 4
\[\therefore \] Probability $=\dfrac{4+4+4}{52}=\dfrac{12}{52}=\dfrac{3}{13}$
(iv) neither an ace nor a king:
The required cards are all cards other than kings and aces.
$\Rightarrow \left( T \right)-\left( number\ of\ kings+number\ of\ aces \right)$
Number of kings = 4
Number of aces = 4
\[\therefore \] Probability $=\dfrac{\left( T \right)-\left( 4+4 \right)}{52}=\dfrac{52-8}{52}=\dfrac{44}{52}=\dfrac{11}{13}$
(v) a spade or an ace
Number of cards of spades = 13
Aces other than spades = 4 – 1 = 3
\[\therefore \] Probability $=\dfrac{3}{13}$
(vi) neither a red card nor a queen
Required cards are all cards other than red cards and queens.
Red cards = 13 x 2 = 26
Queens other than those included in red cards = 4 – 2 = 2
Therefore, probability $=\dfrac{T-\left( 26+2 \right)}{52}$
$\begin{align}
  & =\dfrac{52-28}{52} \\
 & =\dfrac{24}{52} \\
 & =\dfrac{6}{13} \\
\end{align}$
(vii) other than an ace
Required cards are all cards other than ace.
$\Rightarrow T-\left( number\ of\ aces \right)$
Number of aces = 4
\[\therefore \] Probability $=\dfrac{T-4}{52}=\dfrac{52-4}{52}=\dfrac{48}{52}=\dfrac{12}{13}$
(viii) a ten
Number of tens = 4
\[\therefore \] Probability $=\dfrac{4}{52}=\dfrac{1}{13}$
(ix) a spade
Number of spades = 13
\[\therefore \] Probability $=\dfrac{13}{52}=\dfrac{1}{4}$
(x) a black card
Number of black cards = number of spades + number of clubs
= 13 + 13 = 26
\[\therefore \] Probability $=\dfrac{26}{52}=\dfrac{1}{2}$
(xi) the seven of clubs
There are only one seven clubs in one deck of cards.
\[\therefore \] Probability $=\dfrac{1}{52}$
(xii) a jack
Number of jacks = 4
\[\therefore \] Probability $=\dfrac{4}{52}=\dfrac{1}{13}$
(xiii) the ace of spades
There is only one ace of spades in a deck of cards.
\[\therefore \] Probability $=\dfrac{1}{52}$
(xiv) a queen
Number of queens = 4
\[\therefore \] Probability $=\dfrac{4}{52}=\dfrac{1}{13}$
(xv) a heart
Number of hearts = 13
\[\therefore \] Probability $=\dfrac{13}{52}=\dfrac{1}{4}$
(xvi) a red card
Number of red cards = number of hearts + number of diamonds
= 13 + 13 =26
\[\therefore \] Probability $=\dfrac{26}{52}$
(xvii) neither a king nor a queen
The required cards are all cards except kings and queens.
$\begin{align}
  & \Rightarrow T-\left( number\ of\ kings\ +\ number\ of\ queens \right) \\
 & \Rightarrow T-\left( 4+4 \right) \\
\end{align}$
\[\therefore \] Probability $=\dfrac{52-8}{52}=\dfrac{44}{52}=\dfrac{11}{13}$

Note: (1) Make sure to not count a card twice like in part (ii) or (v).
(2) Ace is not a face card. Many students make that mistake.
(3) This is the distribution of a deck of playing cards:
In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each; i.e. spades, hearts, diamonds and clubs. Cards of spades and clubs are black cards. Cards of hearts and diamonds are red cards. The cards in each suit are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.