Show
Questions about how to figure out the probability of picking from a deck of cards common in basic stats courses. For example, the probability of choosing one card, and getting a certain number card (e.g. a 7) or one from a certain suit (e.g. a club). Watch the video for examples: Probability of getting cards from a deck Watch this video on YouTube. Can’t see the video? Click here. You might wonder why you’re learning about cards (what’s the point?). The answer is that finding probabilities (like the probability of contracting an illness) can be a tricky concept to grasp at first. So your instructor will try and simplify problems using cards, dice or Bingo numbers. Once you’ve grasped the basics, you’ll start to use “real life” data for probability (usually a bit later on in the class, for example in normal distributions). Probability of picking from a deck of cards: Steps
That’s it! Tip: It isn’t as easy as just adding the number of sevens (4) and the number of clubs (13). If you did this for this example, you’d get 17 cards, not the correct answer of 16. The reason for this is that one of the cards in our example is both a club AND a number 7. Probability of picking from a deck of cards: Using ExcelWatch the video for an overview and examples of using the hypergeometric distribution in Excel for card probabilities: Excel Hypergeometric distribution to calculate card probabilities Watch this video on YouTube. Can’t see the video? Click here. It gets a LOT more complex if you’re playing a card game, you have a certain number of cards in your hand, and you want to know your odds of getting a certain card if you are drawing a certain number of cards. You have to use something called a hypergeometric distribution to figure out the odds. The formula is: H (n) = C (X, n) * C (Y – X, Z – n) / C (Y, Z) Where: X is the number of a certain card in the deck Y is the total number of cards in the deck Z is the number of cards drawn N is the number you are checking for As you can see, the formula uses combinations and factorials —it can get a bit messy to do this by hand, so consider using technology like Excel. The command in Excel is: “=HYPGEOMDIST(N,Z,X,Y)”. For example, if you have a standard 52 card deck and draw 4 cards, what will be your chances of not drawing an ace? X is 4 Y is 52 Z is 4 N is 0 (as you want zero aces!) the formula would be:
ReferencesBeyer, W. H. CRC Standard Mathematical Tables, 31st ed. Boca Raton, FL: CRC Press, pp. 536 and 571, 2002.
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! Comments? Need to post a correction? Please Contact Us.
In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.
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 outcomesP(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 outcomesP(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 outcomesP(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 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 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 outcomesP(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’ 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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(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 outcomesP(L) = Total number of possible outcome = 38/52 = 19/26 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 randomly picking a Jack from a deck of 52 playing cards?
Pos Terkait
Copyright © 2024 ketiadaan Inc.
