If a Happened What Is the Chance a Happens Again
| Basic Concepts David M. LanePrerequisites Introduction to ProbabilityLearning Objectives
Probability of a Single Event If you roll a half-dozen-sided dice, there are six possible outcomes, and each of these outcomes is equally likely. A half-dozen is every bit probable to come equally a three, and also for the other four sides of the dice. What, and so, is the probability that a one will come up up? Since in that location are six possible outcomes, the probability is one/vi. What is the probability that either a ane or a six will come up? The two outcomes nigh which nosotros are concerned (a 1 or a vi coming upwards) are called favorable outcomes. Given that all outcomes are equally likely, we can compute the probability of a one or a six using the formula: In this case there are two favorable outcomes and six possible outcomes. So the probability of throwing either a one or 6 is 1/iii. Don't be misled by our use of the term "favorable," by the manner. You should understand it in the sense of "favorable to the event in question happening." That outcome might not be favorable to your well-being. You might be betting on a 3, for example. The above formula applies to many games of chance. For example, what is the probability that a carte drawn at random from a deck of playing cards will exist an ace? Since the deck has four aces, at that place are four favorable outcomes; since the deck has 52 cards, at that place are 52 possible outcomes. The probability is therefore 4/52 = one/xiii. What about the probability that the card will exist a social club? Since in that location are 13 clubs, the probability is 13/52 = one/iv. Permit's say you accept a handbag with 20 cherries: 14 sweetness and 6 sour. If you pick a red at random, what is the probability that it volition be sugariness? In that location are twenty possible cherries that could be picked, so the number of possible outcomes is twenty. Of these 20 possible outcomes, xiv are favorable (sweet), so the probability that the cherry will be sugariness is 14/20 = 7/x. At that place is ane potential complication to this case, however. It must be assumed that the probability of picking any of the cherries is the aforementioned as the probability of picking any other. This wouldn't exist truthful if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to paw more readily when you sampled from the pocketbook.) Permit usa keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, nosotros rely heavily on the assumption of equal probability for all outcomes. Here is a more complex example. You throw 2 dice. What is the probability that the sum of the 2 dice will be half dozen? To solve this problem, list all the possible outcomes. There are 36 of them since each die tin come up one of six ways. The 36 possibilities are shown below.
If yous know the probability of an event occurring, it is easy to compute the probability that the event does not occur. If P(A) is the probability of Effect A, then ane - P(A) is the probability that the issue does not occur. For the last instance, the probability that the total is 6 is 5/36. Therefore, the probability that the full is not vi is 1 - 5/36 = 31/36. Probability of Two (or more) Independent Events Events A and B are contained events if the probability of Event B occurring is the aforementioned whether or not Event A occurs. Let'south take a uncomplicated example. A fair coin is tossed ii times. The probability that a caput comes up on the second toss is 1/ii regardless of whether or not a head came up on the get-go toss. The two events are (1) start toss is a caput and (2) second toss is a head. Then these events are independent. Consider the two events (1) "It will rain tomorrow in Houston" and (2) "Information technology will pelting tomorrow in Galveston" (a city well-nigh Houston). These events are non independent because it is more likely that it will pelting in Galveston on days it rains in Houston than on days it does non. Probability of A and B When two events are independent, the probability of both occurring is the production of the probabilities of the individual events. More than formally, if events A and B are independent, then the probability of both A and B occurring is: P(A and B) = P(A) 10 P(B) where P(A and B) is the probability of events A and B both occurring, P(A) is the probability of issue A occurring, and P(B) is the probability of event B occurring. If you flip a money twice, what is the probability that it will come up up heads both times? Effect A is that the coin comes upward heads on the commencement flip and Event B is that the coin comes up heads on the 2nd flip. Since both P(A) and P(B) equal one/2, the probability that both events occur is 1/2 x 1/two = one/4. Permit's take another example. If y'all flip a coin and roll a vi-sided dice, what is the probability that the money comes up heads and the die comes upwards 1? Since the 2 events are independent, the probability is but the probability of a head (which is 1/2) times the probability of the dice coming upwardly 1 (which is ane/6). Therefore, the probability of both events occurring is i/ii ten 1/6 = 1/12. Ane final example: Yous describe a card from a deck of cards, put information technology back, and and so draw another card. What is the probability that the first card is a heart and the second card is black? Since in that location are 52 cards in a deck and 13 of them are hearts, the probability that the showtime card is a heart is 13/52 = 1/4. Since there are 26 black cards in the deck, the probability that the 2d carte du jour is black is 26/52 = 1/ii. The probability of both events occurring is therefore i/four 10 1/2 = 1/eight. Encounter the section on conditional probabilities on this page to run into how to compute P(A and B) when A and B are not independent. Probability of A or B If Events A and B are contained, the probability that either Event A or Effect B occurs is: P(A or B) = P(A) + P(B) - P(A and B) In this discussion, when nosotros say "A or B occurs" we include three possibilities:
This use of the discussion "or" is technically called inclusive or considering it includes the case in which both A and B occur. If we included only the first two cases, then nosotros would be using an exclusive or. (Optional) Nosotros tin can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:
The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the unproblematic event B happens if either A-and-B happens or non-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-non-B) + P(non-A-and-B). Nosotros tin can make these two sums equal past subtracting i occurrence of P(A-and-B) from the starting time. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B). Now for some examples. If you flip a money 2 times, what is the probability that y'all will go a head on the outset flip or a head on the second flip (or both)? Letting Event A be a head on the offset flip and Issue B be a head on the second flip, so P(A) = one/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore, P(A or B) = 1/ii + ane/2 - i/4 = 3/4. If y'all throw a six-sided die and so flip a coin, what is the probability that you will get either a half-dozen on the die or a caput on the coin flip (or both)? Using the formula, P(6 or caput) = P(6) + P(head) - P(half-dozen and head) An alternating approach to calculating this value is to commencement by calculating the probability of not getting either a 6 or a head. Then decrease this value from 1 to compute the probability of getting a vi or a head. Although this is a complicated method, it has the reward of being applicable to bug with more than two events. Hither is the adding in the present case. The probability of non getting either a 6 or a head tin be recast equally the probability of (non getting a 6) AND (not getting a head). This follows because if you did not get a 6 and you did not get a head, so yous did not go a 6 or a head. The probability of not getting a vi is 1 - i/6 = v/6. The probability of non getting a head is 1 - i/2 = 1/2. The probability of not getting a six and non getting a caput is 5/6 x ane/2 = v/12. This is therefore the probability of non getting a six or a head. The probability of getting a half dozen or a head is therefore (once once again) 1 - 5/12 = 7/12. If y'all throw a dice three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a one on the 2d throw OR a 1 on the third throw? The easiest manner to approach this problem is to compute the probability of NOT getting a 1 on the first throw The answer will be 1 minus this probability. The probability of not getting a one on whatever of the three throws is 5/half-dozen 10 v/6 ten 5/half dozen = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is i - 125/216 = 91/216. Conditional Probabilities Often it is required to compute the probability of an upshot given that some other event has occurred. For example, what is the probability that ii cards drawn at random from a deck of playing cards will both be aces? It might seem that you could use the formula for the probability of 2 independent events and only multiply four/52 10 iv/52 = 1/169. This would be wrong, however, considering the 2 events are non independent. If the kickoff card drawn is an ace, and so the probability that the second card is as well an ace would be lower because in that location would but be three aces left in the deck. Once the first bill of fare chosen is an ace, the probability that the 2nd card chosen is also an ace is called the conditional probability of drawing an ace. In this example, the "status" is that the first card is an ace. Symbolically, we write this as: P(ace on second draw | an ace on the get-go draw) The vertical bar "|" is read equally "given," and so the above expression is short for: "The probability that an ace is drawn on the 2d draw given that an ace was drawn on the outset draw." What is this probability? Since after an ace is drawn on the outset draw, in that location are 3 aces out of 51 total cards left. This means that the probability that ane of these aces volition be drawn is 3/51 = 1/17. If Events A and B are not independent, then P(A and B) = P(A) x P(B|A). Applying this to the problem of ii aces, the probability of drawing two aces from a deck is 4/52 x 3/51 = i/221. One more than example: If yous depict ii cards from a deck, what is the probability that yous will go the Ace of Diamonds and a black card? There are two means you lot can satisfy this condition: (1) You can get the Ace of Diamonds first and so a blackness card or (2) you tin get a black card first and and then the Ace of Diamonds. Let's calculate Case A. The probability that the start menu is the Ace of Diamonds is ane/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 considering 26 of the remaining 51 cards are black. The probability is therefore ane/52 x 26/51 = 1/102. Now for Case 2: the probability that the first carte is black is 26/52 = i/2. The probability that the second card is the Ace of Diamonds given that the showtime card is black is ane/51. The probability of Example 2 is therefore 1/ii ten ane/51 = ane/102, the aforementioned as the probability of Instance 1. Call back that the probability of A or B is P(A) + P(B) - P(A and B). In this problem, P(A and B) = 0 since a carte cannot be the Ace of Diamonds and be a blackness card. Therefore, the probability of Case 1 or Case 2 is 1/102 + 1/102 = 2/102 = one/51. So, one/51 is the probability that you lot will get the Ace of Diamonds and a black card when drawing ii cards from a deck. Birthday Problem If there are 25 people in a room, what is the probability that at least 2 of them share the aforementioned birthday. If your beginning thought is that it is 25/365 = 0.068, you will be surprised to learn information technology is much college than that. This problem requires the application of the sections on P(A and B) and conditional probability. This problem is best approached by asking what is the probability that no 2 people have the same birthday. Once we know this probability, we can simply subtract it from 1 to find the probability that two people share a birthday. If we choose ii people at random, what is the probability that they do non share a birthday? Of the 365 days on which the second person could have a birthday, 364 of them are unlike from the first person's altogether. Therefore the probability is 364/365. Let'south define P2 equally the probability that the second person drawn does not share a altogether with the person drawn previously. P2 is therefore 364/365. Now define P3 as the probability that the third person drawn does not share a altogether with anyone drawn previously given that there are no previous birthday matches. P3 is therefore a conditional probability. If there are no previous altogether matches, and so two of the 365 days accept been "used up," leaving 363 non-matching days. Therefore P3 = 363/365. In like fashion, P4 = 362/365, P5 = 361/365, and so on upwardly to P25 = 341/365. In order for in that location to be no matches, the second person must non match any previous person and the third person must not match any previous person, and the fourth person must not lucifer any previous person, etc. Since P(A and B) = P(A)P(B), all we have to do is multiply P2, P3, P4 ...P25 together. The result is 0.431. Therefore the probability of at to the lowest degree one match is 0.569. Gambler'due south Fallacy A fair coin is flipped five times and comes upwardly heads each time. What is the probability that it volition come up heads on the sixth flip? The correct answer is, of course, 1/2. Merely many people believe that a tail is more than probable to occur after throwing five heads. Their faulty reasoning may get something similar this: "In the long run, the number of heads and tails volition be the same, so the tails have some catching up to practise." The flaws in this logic are exposed in the simulation in this chapter. Please reply the questions: |
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