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## Probability, Part III

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**Learning Objectives**By the end of this lecture, you should be able to: • Describe what is meant by independent and non-independent events. • Describe and apply the multiplication rule for independent events. • Describe two common ways of coming up with probabilities. • Compare and contrast the concepts of disjointness v.s. independence.**Methods of determiningthe probability to an event:**There are 2 well-established ways of assigning probabilities with some degree of accuracy: • Theoretically from our understanding of the phenomenon and symmetries in the problem • For a 6-sided fair die, P(2) = 1/6 • For a deck of cards, P(Ace of Hearts) = 1/52 • Empirically We look at data. In other words, we are deriving our knowledge based on numerous similar past events. We determine a probability empirically when there isn’t a theoretical method of determining the probability. • EXAMPLE: What is the probability that a random student in IT-223 will receive an A? This probability can not be determine theoretically the way, say, a coin flip can be. Instead, we’d have to look at past data from IT-223 courses and see how many As were recorded. If in a sample of 600 students, 54 A’s were recorded, then P(grade of A) = 54 / 600. • EXAMPLE: What is the probability of a baseball player getting a hit on a given at bat? Again, we would have to look at past data to see his/her batting average. If over the last year, he hit 122 times out of 411 at bats, P(Hit) = 122/400.**An easy way of determining a probability when all outcomes**are equally likely: When all outcomes are equally likely (have the same probability), there is a nice quick way of calculating probabilities: If all of the outcomes are equally likely (i.e. they all have the same probabilities), such as a die roll or a coin flip, then each individual outcome has probability: 1 / (# of possible outcomes). • E.g.: For a coin flip, all possible outcomes have an equal probability of 1 / (2 possible outcomes) = 0.5 • E.g.: For a die roll, all outcomes have an equal probability of 1 / (6 possible outcomes) = 0.167 Formula: For any event A where all outcomes have the same probability:**Dice**You toss two dice. What is the probability of the outcomes summing to 5? S: {(1,1), (1,2), (1,3), ……etc.} • There are 36 possible outcomes in S, all equally likely. • Thus, the probability of any one of them is 1/36. • Events having a sum of 5 has 4 possible outcomes (1&4, 2&3, 4&1, 3&2) • 36 outcomes in sample space • = 4/36**EXAMPLE: What is the probability of the ball landing in an**even number slot? Answer: All slots have an equal probability. So: P(Even) = count of Even outcomes / count of All outcomes = 18 / 38 = 0.47**Probability rules (cont’d)**#5: Multiplication Rule for Independent Events The probability of one event occurring AND another event occurring equals the probability of event 1 times the probability of event 2. However, this rule only holds if the events are independent. Two events A and B are independent if knowing that one event has occurred does not change the probability that the other will occur.**Two events are independent if the probability that one event**occurs on one trial of an experiment is not affected or changed by the occurrence of any other event (e.g. two consecutive coin tosses) . When are events NOT independent? Imagine that these coins were spread out so that half the coins showed heads and half showed tails. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin that is heads up is now less than 0.5. If you were to put the coin back each time, then the trials would be considered independent!**Are the events independent?**• Flip a coin once and get ‘heads’. Does this result change the probability of rolling a heads again on the second roll? • Event 1: Get a heads on first flip. Event 2: Get a heads on second flip. The probability of event #2 is NOT affected by whatever happened on event #1. Therefore, these are independent events • You buy a lottery ticket and win. Then you win again the next week. Does this change the likelihood of winning on the third week since you are “on a roll”? • Event 1: Win the first week. Event 2: Win the second week. Event 3: Win the third week. However, winning the first two weeks in no way affects the likelihood of winning the 3rd week. In other words, these events are all independent • In a game of Poker, you draw a card from the deck and get an Ace. Does this result change the probability of getting an Ace on a second draw from that same deck? • Event 1: First card is an Ace. Event 2: Second card is an Ace. The probability of event #2 IS affected by whatever happened in event #1. Therefore, these events are NOT independent. (Initially, 4/52 chance, after removing 1 ace, 3/51 chance)**Probability rules (cont’d)**#5: Multiplication Rule for Independent Events P(A and B) = P(A) * P(B) Again, this rule only applies to independent events. Later we will show a second rule that applies to non-independent events**“AND” Questions**• Recall how in a previous discussion, we referred to “or” questions. • E.g. I need to decide what to wear to work: What is the probability that it will be snowing or it will be raining? • Another extremely common real world scenario involves “AND” questions: • I need to roll a 5 and a 2 in a game of backgammon to win the game. • I need to draw two cards in which one must be an Ace and the other must be a 7. • If the Cubs win and the Brewers lose, the Cubs win the division title. For “and” questions, we use the multiplication rule. Though only if the events are independent!!**Example**What is the probability of getting a tails on two consecutive coin tosses? We are asking the probability of getting a tails on the first coin toss and on the second coin toss. With ‘and’ questions, we typically think of the multiplication rule. At that point we must ask ourselves if the events are independent. Because we are discussing two consecutive coin tosses, which means that whatever happens in the first coin toss does not affect what happens on the second coin toss, the events are indeed independent. Therefore, we may use our multiplication rule for independent events: P(first = Tail and second = Tail) = P(first Tail) * P(second Tail) = 0.5 * 0.5 = 0.25**Example:**A couple intends to have three children. What is the likelihood that they will have only boys? Answer: After a little bit of thought, you realize that you are being asked P(first child is a boy) AND P(2nd child is a boy) AND P(3rd child is a boy). ‘AND’ questions should make you think of the multiplication rule. At that point, you need to ask yourself if these 3 events are independent. In fact, they are. So we may therefore use the multiplication rule: P(BBB) = P(B)* P(B)* P(B) = (1/2)*(1/2)*(1/2) = 1/8**Example**• You are dealt two cards out of a deck. Calculate the probability that the cards are the Ace of Spades and the Ace of Hearts. • Solution: You will probably be tempted to use the multiplication rule. However, this rule (in its current form) only applies to independent events. As it turns out, these events are not independent. If one of the events has occurred (e.g. drawing the ace of spades from the deck), the probability of the second event changes as a result. • Later, we will learn a slight modification to this rule so that it can also be applied to non-independent events. • If, however, we changed the question to state that instead of pulling two cards out of the deck, we pull out one card, then put it back and pull out another card, we could use our multiplication rule, since these are now independent events.**Example**• A state lottery’s Pick 3 game asks players to choose a 3-digit number, 000 to 999. The state chooses the winning number at random, so each number has probability 1/1000. You win if the winning number contains the digits in your number, in any order. • Example #1: Your number is 456. What is your probability of winning? (Write on a piece of paper) • Example #2: Your number is 212. What is your probability of winning. • Solution: As always, you have to read the question carefully, and always be sure to watch the fine print! In this case, the three tiny words at the end, “in any order” change the entire nature of the problem. • So, instead of winning with 456 where your probability would be 1/1000, you can also win with 465, or 546, or 564 etc. If you look at all possible permutations, you will see that there are 6 distinct possibilities: {456, 465, 546, 564, 645, 654}. Since each one has a 1/1000 chance of winning, and you have six possibilities, the probability of winning is 6/1000 or 0.006. • For the second part, you only have three distinct arrangements: {212, 221, 122}. So your probability of winning is 1/1000 + 1/1000 + 1/1000 = 0.003. Incidentally, did you try to use the multiplication rule here? In fact, you should be using the addition rule! Remember that you must always make sure you are using the right tool for the job!!**Example**• The PINs for ATMs usually consist of 4 digits. You notice that most PINs have at least one 0 and you wonder if the issuers use lots of 0s to make the numbers easy to remember. Suppose that PINs are assigned at random so that all 4-digit numbers are equally likely. • How many possible PINs are there? – Write down your answer • What is the probability that a PIN assigned at random has at least one 0? – Write. SOLUTION: • Possible PINs: 104 = 10,000 • The operative word for the second question is “at least one”. In other words, what is the probability that 1 or 2 or 3 or all 4 numbers are zero? This turns out to be a surprisingly involved calculation. However, if you remember your trusty complement rule, it becomes surprisingly easy! You realize that you could phrase things another way: If I can determine the probability that NONE of the numbers are 0, then the complement of that equals the probability AT LEAST one is 0. • P(one number = 0) = 0.1, P(not 0)=0.9 • P(all 4 numbers are not 0) = 0.9*0.9*0.9*0.9 = 0.6561. Don’t forget that this is the probability that NONE are 0. So the probability that 1 or more are zero = 1-0.6561 = 0.3439.**Disjoint vs Independent**• The two are often confused. Be sure that you are clear on the meaning of each. • One common question has to do with the relationship between them. Is there one? For example, can an event be both disjoint and independent? • Disjoint = If two events, A and B are disjoint, we are saying that if the event ‘A’ A is true, then P(B) can not be true. So let’s say that our event is indeed disjoint. • If our event is disjoint, this says that P(B) is affected by A. Well, by definition, if the probability of B (i.e. P(B) ) is indeed changed depending on whether or not A occurs , then P(B) is indeed DE-pendent on A! • So the answer to our question is NO! When you think about it, if two events are disjoint, then those events can NOT be independent. I fully recognize that you may need to spend some time wrapping your brain around this. That’s okay – however, it is something that you should do. Think of it as ‘brain exercise’! (Plus it helps to make sure you understand the concept).