Return to Statistics Topics Summary of Basic ProbabilityThe classical or theoretical definition of probability assumes that there are a finite number of outcomes in a situation and all the outcomes are equally likely. Classical Definition of Probability
Though you probably have not seen this definition before, you probably have an inherent grasp of the concept. In other words, you could guess the probabilities without knowing the definition. Cards and Dice The examples that follow require some knowledge of cards and dice. Here are the basic facts needed compute probabilities concerning cards and dice. A standard deck of cards has four suites: hearts, clubs, spades, diamonds. Each suite has thirteen cards: ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen and king. Thus the entire deck has 52 cards total. When you are asked about the probability of choosing a certain card from a deck of cards, you assume that the cards have been well-shuffled, and that each card in the deck is visible, though face down, so you do not know what the suite or value of the card is. A pair of dice consists of two cubes with dots on each side. One of the cubes is called a die, and each die has six sides.Each side of a die has a number of dots (1, 2, 3, 4, 5 or 6), and each number of dots appears only once. Example 1 The probability of choosing a heart from a deck of cards is given by
Example 2 The probability of choosing a three from a deck of cards is
Example 3 The probability of a two coming up after rolling a die (singular for dice) is
The classical definition works well in determining probabilities for games of chance like poker or roulette, because the stated assumptions readily apply in these cases. Unfortunately, if you wanted to find the probability of something like rain tomorrow or of a licensed driver in Louisiana being involved in an auto accident this year, the classical definition does not apply. Fortunately, there is another definition of probability to apply in these cases. Empirical Definition of Probability The probability of event A is the number approached by
as the total number of recorded outcomes becomes "very large." The idea that the fraction in the previous definition will approach a certain number as the total number of recorded outcomes becomes very large is called the Law of Large Numbers. Because of this law, when the Classical Definition applies to an event A, the probabilities found by either definition should be the same. In other words, if you keep rolling a die, the ratio of the total number of twos to the total number of rolls should approach one-sixth. Similarly, if you draw a card, record its number, return the card, shuffle the deck, and repeat the process; as the number of repetitions increases, the total number of threes over the total number of repetitions should approach 1/13 ≈ 0.0769. In working with the empirical definition, most of the time you have to settle for an estimate of the probability involved. This estimate is thus called an empirical estimate. Example 4 To estimate the probability of a licensed driver in Louisiana being involved in an auto accident this year, you could use the ratio
To do better than that, you could use the number of accidents for the last five years and the total number of Louisiana drivers in the last five years. Or to do even better, use the numbers for the last ten years or, better yet, the last twenty years. Example 5 Estimating the probability of rain tomorrow would be a little more difficult. You could note today's temperature, barometric pressure, prevailing wind direction, and whether or not there are rain clouds that could be blown into your area by tomorrow. Then you could find all days on record in the past with similar temperatures, pressures, and wind directions, and clouds in the right location. Your rainfall estimate would then be the ratio
To make your estimate better, you might want to add in humidity, wind speed, or season of the year. Or maybe if there seemed to be no relation between humidity levels and rainfall, you might want add in the days that did not meet your humidity level requirements and thus increase the total number of days. Example 6 If you want to estimate the probability that a dam will burst, or a bridge will collapse, or a skyscraper will topple, there is usually not much past data available. The next best thing is to do a computer simulation. Simulation results can be compiled a lot faster with a lot less money and less loss of life than actual events. The estimated probability of say a bridge collapsing would be given by the following fraction
The more true to life the simulation is, the better the estimate will be. Basic Probability Rules For either definition, the probability of an event A is always a number between zero and one, inclusive; i.e.
Sometimes probability values are written using percentages, in which case the rule just given is written as follows
If the event A is not possible, then P(A) = 0 or P(A) = 0%. If event A is certain to occur, then P(A) = 1 or P(A) = 100%. The sum of the probabilities for each possible outcome of an experiment is 1 or 100%. This is written mathematically as follows using the capital Greek letter sigma (S) to denote summation.
Probability Scale* The best way to find out what the probability of an event means is to compute the probability of a number of events you are familiar with and consider how the probabilities you compute correspond to how frequently the events occur. Until you have computed a large number of probabilities and developed your own sense of what probabilities mean, you can use the following probability scale as a rough starting point. When you gain more experience with probabilities, you may want to change some terminology or move the boundaries of the different regions.
*This is a revised and expanded version of the probability scale presented in Mario Triola, Elementary Statistics Using the Graphing Calculator: For the TI-83/84 Plus, Pearson Education, Inc. 2005, page 135. Return to Statistics Topics
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’ Number of favorable outcomesP(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’ 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 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.
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What is the probability that a single card drawn from a standard deck of 52 cards will be a jack or a heart?
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