Playing cards probability problems based on a well-shuffled deck of 52 cards. Show 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 = 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 = 4/52 (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 = 2/52 (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
= 13/52 (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 = 8/52 (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 = 40/52 (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 = 6/52 (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 = 26/52 (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 = 48/52 (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
= 20/52 (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 = 36/52 (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 = 38/52 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. What is the probability of getting a 10 of clubs?2 What is the probability of drawing a 10 of clubs at random from a deck of 52 cards? There is only one 10 Of clubs in a deck of 52 cards, so the probably would be 1/52.
What is the probability of getting a 10 or a jack in a deck of cards?The probability of both outcomes is equal i.e. 50% or 1/2. So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event). What is Sample Space?
What is the probability of getting a club card?For example, what is the probability of drawing a card from a standard deck of cards and getting a face card or a club? The probability of drawing a club is 13/52. The probability of drawing a face card is 12/52.
What is the probability that the card drawn is 10 or spade?∴ Probability of drawing a 10 or a spade = P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 4 52 + 13 52 − 1 52 = 16 52 = 4 13 .
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