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

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Cards of Spades and also clubs are black cards.

Cards that hearts and diamonds room red cards.

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

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

Worked-out troubles on playing cards probability:

1. A card is drawn from a fine shuffled load of 52 cards. Uncover theprobability of:

(i) ‘2’ of spades

(ii) a jack

(iii) a king of red colour

(iv) a card of diamond

(v) a king or a queen

(vi) a non-face card

(vii) a black confront card

(viii) a black card

(ix) a non-ace

(x) non-face map of black color colour

(xi) neither a spade no one a jack

(xii) no 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’ of spades:

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

Therefore, probability of acquiring ‘2’ ofspade

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

(ii) a jack

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

Therefore, probability of acquiring ‘a jack’

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

(iii) a king that red colour

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

Therefore, probability of getting ‘a kingof red colour’

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

(iv) a map of diamond

Number of 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 the end of 52 cards.

Total number of queen is 4 the end of 52 cards

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

Therefore, probability of gaining ‘a kingor a queen’

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

(vi) a non-face card

Total number of face card out of 52 cards =3 time 4 = 12

Total number 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 outcome = 40/52 = 10/13

(vii) a black confront card:

Cardsof Spades and also Clubs room black cards.

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

number of face map in clubs (king, queen andjack or knaves) = 3

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

Therefore, probability of getting ‘a blackface card’

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

(viii) a black color card:

Cards that spades and clubs are black cards.

Number the spades = 13

variety of clubs = 13

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

Therefore, probability of obtaining ‘a blackcard’

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

(ix) a non-ace:

Number the 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 variety of non-ace cards out of52 cards = 52 - 4

= 48

Therefore, probability of obtaining ‘anon-ace’

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

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

Cards of spades and clubs are black cards.

Number the spades = 13

variety of clubs = 13

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

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

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

Therefore, probability of obtaining ‘non-facecard of black colour’

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

(xi) no a spade no one a jack

Number that spades = 13

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

Number of jack the end of 52 cards = 4

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

Neither a spade nor a jack = 39 - 3 = 36

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

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

(xii) neither a heart no one a red king

Number of understanding = 13

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

Therefore, spades, clubs and also diamonds arethe 39 cards.

Cardsof hearts and diamonds are red cards.

Number of red monarchs in red cards = 2

Therefore, neither a heart no one a red king =39 - 1 = 38

Therefore, probability of obtaining ‘neithera heart no one a red king’

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

2. A card is attracted at random from a well-shuffled pack of cards numbered 1 come 20. Discover the probability of

(i) acquiring a number less than 7

(ii) getting a number divisible by 3.

Solution:

(i) Total variety of possible outcomes = 20 ( due to the fact that there space cards numbered 1, 2, 3, ..., 20).

Number the favourable outcomes for the occasion E

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

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

= \$$\\frac620\$$

= \$$\\frac310\$$.

(ii) Total number of possible outcomes = 20.

Number that favourable outcomes for the event F

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

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

= \$$\\frac620\$$

= \$$\\frac310\$$.

3. A map is drawn at random from a pack of 52 playing cards. Discover the probability that the card attracted is

(i) a king

(ii) no a queen nor a jack.

Solution:

Total number of possible outcomes = 52 (As there are 52 various cards).

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

So, through definition, P(E) = \$$\\frac452\$$

= \$$\\frac113\$$.

(ii) number of favourable outcomes because that the occasion F

= variety of cards which room neither a queen no one a jack

= 52 - 4 - 4, .

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= 44

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

= \$$\\frac1113\$$.