Playing Cards Probability

Playing cards probability problems based on a well-shuffled deck of 52 cards.


Basic concept on drawing a card:

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts , diamonds , clubs .

Cards of Spades and clubs are black cards.

Cards of hearts and diamonds are red cards.

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

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


Worked-out problems on Playing cards probability:

1. A card is drawn from a well shuffled pack of 52 cards. Find the probability 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 face card

(viii) a black card

(ix) a non-ace

(x) non-face card of black colour

(xi) neither a spade nor a jack

(xii) neither a heart nor a red king

Solution:

In a playing card there are 52 cards.

Therefore the total number of possible outcomes = 52

(i) ‘2’ of spades:

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

Therefore, probability of getting ‘2’ of spade

               Number of favorable outcomes
P(A) =     Total number of possible outcome


      = 1/52

(ii) a jack

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

Therefore, probability of getting ‘a jack’

               Number of favorable outcomes
P(B) =     Total number of possible outcome


      = 4/52
      = 1/13

(iii) a king of red colour

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

Therefore, probability of getting ‘a king of red colour’

               Number of favorable outcomes
P(C) =     Total number of possible outcome


      = 2/52
      = 1/26

(iv) a card of diamond

Number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards.

Therefore, probability of getting ‘a card of diamond’

               Number of favorable outcomes
P(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 number of queen is 4 out of 52 cards

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

Therefore, probability of getting ‘a king or a queen’

               Number of favorable outcomes
P(E) =     Total number of possible outcome


      = 8/52
      = 2/13

(vi) a non-face card

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

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

Therefore, probability of getting ‘a non-face card’

               Number of favorable outcomes
P(F) =     Total number of possible outcome


      = 40/52
      = 10/13

(vii) a black face card:

Cards of Spades and Clubs are black cards.

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

Number of face card in clubs (king, queen and jack or knaves) = 3

Therefore, total number of black face card out of 52 cards = 3 + 3 = 6

Therefore, probability of getting ‘a black face card’

               Number of favorable outcomes
P(G) =     Total number of possible outcome


      = 6/52
      = 3/26

(viii) a black card:

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

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

Therefore, probability of getting ‘a black card’

               Number of favorable outcomes
P(H) =     Total number of possible outcome


      = 26/52
      = 1/2

(ix) a non-ace:

Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = 1

Therefore, total number of ace cards out of 52 cards = 4

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

= 48

Therefore, probability of getting ‘a non-ace’

               Number of favorable outcomes
P(I) =     Total number of possible outcome


      = 48/52
      = 12/13

(x) non-face card of black colour:

Cards of spades and clubs are black cards.

Number of spades = 13

Number of clubs = 13

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

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

Therefore, total number of non-face card of black colour out of 52 cards = 26 - 6 = 20

Therefore, probability of getting ‘non-face card of black colour’

               Number of favorable outcomes
P(J) =     Total number of possible outcome


      = 20/52
      = 5/13

(xi) neither a spade nor a jack

Number of spades = 13

Total number of non-spades out of 52 cards = 52 - 13 = 39

Number of jack out of 52 cards = 4

Number of jack in each of three suits namely hearts, diamonds and clubs = 3

[Since, 1 jack is already included in the 13 spades so, here we will take number of jacks is 3]

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

Therefore, probability of getting ‘neither a spade nor a jack’

               Number of favorable outcomes
P(K) =     Total number of possible outcome


      = 36/52
      = 9/13

(xii) neither a heart nor a red king

Number of hearts = 13

Total number of non-hearts out of 52 cards = 52 - 13 = 39

Therefore, spades, clubs and diamonds are the 39 cards.

Cards of hearts and diamonds are red cards.

Number of red kings in red cards = 2

Therefore, neither a heart nor a red king = 39 - 1 = 38

[Since, 1 red king is already included in the 13 hearts so, here we will take number of red kings is 1]

Therefore, probability of getting ‘neither a heart nor a red king’

               Number of favorable outcomes
P(L) =     Total number of possible outcome


      = 38/52
      = 19/26

Playing Cards Probability


2. A card is drawn at random from a well-shuffled pack of cards numbered 1 to 20. Find the probability of

(i) getting a number less than 7

(ii) getting a number divisible by 3.

Solution:

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

Number of favourable outcomes for the event E

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

So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{6}{20}\)

             = \(\frac{3}{10}\).


(ii) Total number of possible outcomes = 20.

Number of favourable outcomes for the event F

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

So, P(F) = \(\frac{\textrm{Number of Favourable Outcomes for the Event F}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{6}{20}\)

             = \(\frac{3}{10}\).


3. A card is drawn at random from a pack of 52 playing cards. Find 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 are 52 different cards).

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

So, by definition, P(E) = \(\frac{4}{52}\)

                                 = \(\frac{1}{13}\).


(ii) Number of favourable outcomes for the event F

                    = number of cards which are neither a queen nor a jack

                    = 52 - 4 - 4, [Since there are 4 queens and 4 jacks].

                    = 44

Therefore, by definition, P(F) = \(\frac{44}{52}\)

                                          = \(\frac{11}{13}\).

These are the basic problems on probability with playing cards.

Probability

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice









9th Grade Math

From Playing Cards Probability to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Divide by Repeated Subtraction | Division as Repeated Subtraction

    Jan 20, 25 03:34 PM

    Repeated Subtraction
    How to divide by repeated subtraction? We will learn how to find the quotient and remainder by the method of repeated subtraction a division problem may be solved.

    Read More

  2. 3rd Grade Multiplication Worksheet | Grade 3 Multiplication Questions

    Jan 20, 25 02:31 PM

    3rd Grade Multiplication Riddle
    In 3rd Grade Multiplication Worksheet we will solve how to multiply 2-digit number by 1-digit number without regrouping, multiply 2-digit number by 1-digit number with regrouping, multiply 3-digit num…

    Read More

  3. 3rd Grade Math Worksheets |3rd Grade Math Sheets|3rd Grade Math Lesson

    Jan 20, 25 12:28 AM

    3rd Grade Math Worksheets
    3rd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students. Teachers and parents can also follow the worksheets to guide the students.

    Read More

  4. 3rd Grade Multiplication Word Problems Worksheet With Answers | Math

    Jan 19, 25 11:29 PM

    In 3rd Grade Multiplication Word Problems Worksheet we will solve different types of problems on multiplication, multiplication word problems on 3-digits number by 1-digit number and multiplication wo…

    Read More

  5. Multiplying 2-Digit Numbers by 2-Digit Numbers |Multiplying by 2-Digit

    Jan 19, 25 02:34 AM

    Multiplying 2-Digit Numbers by 2-Digit Numbers
    We will learn how to multiply 2-digit numbers by 2-digit numbers.

    Read More