In a fill or deck that 52 play cards, castle are divided into 4 suits of 13 cards each i.e. Spades ♠ hearts ♥, diamonds ♦, clubs ♣.

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Cards of Spades and also clubs are black cards.

Cards the hearts and also diamonds are red cards.


The card in every suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and also 2.

King, Queen and also Jack (or Knaves) are face cards. So, there space 12 confront cards in the deck that 52 play cards.

Worked-out problems on playing cards probability:

1. A map is attracted from a fine shuffled fill of 52 cards. Find theprobability of:

(i) ‘2’ the spades

(ii) a jack

(iii) a king of red colour

(iv) a map of diamond

(v) a king or a queen

(vi) a non-face card

(vii) a black face card

(viii) a black card

(ix) a non-ace

(x) non-face map of black colour

(xi) no a spade no one a jack

(xii) neither a heart no one a red king

Solution:

In a play card there space 52 cards.

Therefore the total variety of possibleoutcomes = 52

(i) ‘2’ the spades:

Number that favourable outcomes i.e. ‘2’ ofspades is 1 out of 52 cards.

Therefore, probability of acquiring ‘2’ ofspade

variety of favorable outcomesP(A) = Total number of possible result = 1/52

(ii) a jack

Number the favourable outcomes i.e. ‘a jack’is 4 the end of 52 cards.

Therefore, probability of gaining ‘a jack’

variety of favorable outcomesP(B) = Total number of possible outcome = 4/52 = 1/13

(iii) a king that red colour

Number that favourable outcomes i.e. ‘a kingof red colour’ is 2 out of 52 cards.

Therefore, probability of gaining ‘a kingof red colour’

number of favorable outcomesP(C) = Total number of possible outcome = 2/52 = 1/26

(iv) a map of diamond

Number the favourable outcomes i.e. ‘a cardof diamond’ is 13 the end of 52 cards.

Therefore, probability of obtaining ‘a cardof diamond’

number of favorable outcomesP(D) = Total number of possible outcome = 13/52 = 1/4

(v) a king or a queen

Total number of king is 4 out of 52 cards.

Total variety of queen is 4 the end of 52 cards

Number of favourable outcomes i.e. ‘a kingor a queen’ is 4 + 4 = 8 the end of 52 cards.

Therefore, probability of getting ‘a kingor a queen’

number of favorable outcomesP(E) = Total number of possible outcome = 8/52 = 2/13

(vi) a non-face card

Total variety of face card the end of 52 cards =3 times 4 = 12

Total variety of non-face card out of 52cards = 52 - 12 = 40

Therefore, probability of getting ‘anon-face card’

number of favorable outcomesP(F) = Total variety of possible result = 40/52 = 10/13

(vii) a black challenge card:

Cardsof Spades and Clubs room black cards.

Number of face card in spades (king, queenand jack or knaves) = 3

variety of face card in clubs (king, queen andjack or knaves) = 3

Therefore, total number of black confront cardout the 52 cards = 3 + 3 = 6

Therefore, probability of acquiring ‘a blackface card’

variety of favorable outcomesP(G) = Total variety of possible outcome = 6/52 = 3/26

(viii) a black color card:

Cards of spades and clubs are black cards.

Number the spades = 13

variety of clubs = 13

Therefore, total number of black card outof 52 cards = 13 + 13 = 26

Therefore, probability of acquiring ‘a blackcard’

variety of favorable outcomesP(H) = Total number of possible outcome = 26/52 = 1/2

(ix) a non-ace:

Number that ace cards in each of four suits namelyspades, hearts, diamonds and also clubs = 1

Therefore, total number of ace cards the end of52 cards = 4

Thus, total number of non-ace cards out of52 cards = 52 - 4

= 48

Therefore, probability of gaining ‘anon-ace’

number of favorable outcomesP(I) = Total number of possible result = 48/52 = 12/13

(x) non-face card of black color colour:

Cards the spades and also clubs are black cards.

Number that spades = 13

number of clubs = 13

Therefore, total variety of black map outof 52 cards = 13 + 13 = 26

Number of face cards in each suits namelyspades and clubs = 3 + 3 = 6

Therefore, total number of non-face card ofblack colour the end of 52 cards = 26 - 6 = 20

Therefore, probability of getting ‘non-facecard of black color colour’

number of favorable outcomesP(J) = Total variety of possible outcome = 20/52 = 5/13

(xi) no a spade nor a jack

Number that spades = 13

Total variety of non-spades the end of 52 cards= 52 - 13 = 39

Number of jack out of 52 cards = 4

Number that jack in every of 3 suitsnamely hearts,diamonds and also clubs = 3

Neither a spade no one a jack = 39 - 3 = 36

Therefore, probability of getting ‘neithera spade no one a jack’

number of favorable outcomesP(K) = Total number of possible result = 36/52 = 9/13

(xii) no a heart no one a red king

Number of mind = 13

Total variety of non-hearts the end of 52 cards= 52 - 13 = 39

Therefore, spades, clubs and diamonds arethe 39 cards.

Cardsof hearts and also diamonds space red cards.

Number that red majesties in red cards = 2

Therefore, no a heart no one a red king =39 - 1 = 38

Therefore, probability of getting ‘neithera heart nor a red king’

variety of favorable outcomesP(L) = Total variety of possible result = 38/52 = 19/26
*

2. A map is attracted at arbitrarily from a well-shuffled pack of cards numbered 1 to 20. Discover the probability of

(i) getting a number much less than 7

(ii) obtaining a number divisible by 3.

Solution:

(i) Total number of possible outcomes = 20 ( since there are cards numbered 1, 2, 3, ..., 20).

Number that favourable outcomes for the occasion E

                                = variety of cards showing much less than 7 = 6 (namely 1, 2, 3, 4, 5, 6).

So, P(E) = \(\frac\textrmNumber that Favourable Outcomes because that the event E\textrmTotal variety of Possible Outcomes\)

             = \(\frac620\)

             = \(\frac310\).

(ii) Total variety of possible outcomes = 20.

Number of favourable outcomes because that the occasion F

                                = variety of cards reflecting a number divisible by 3 = 6 (namely 3, 6, 9, 12, 15, 18).

So, P(F) = \(\frac\textrmNumber the Favourable Outcomes because that the event F\textrmTotal variety of Possible Outcomes\)

             = \(\frac620\)

             = \(\frac310\).

3. A card is drawn at arbitrarily from a load of 52 play cards. Discover the probability that the card drawn is 

(i) a king

(ii) neither a queen nor a jack.

Solution:

Total number of possible outcomes = 52 (As there space 52 different cards).

(i) number of favourable outcomes for the occasion E = number of kings in the fill = 4.

So, by definition, P(E) = \(\frac452\)

                                 = \(\frac113\).

(ii) variety of favourable outcomes for the event F

                    = number of cards which are neither a queen no one a jack

                    = 52 - 4 - 4, .

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                    = 44

Therefore, through definition, P(F) = \(\frac4452\)

                                          = \(\frac1113\).