HYPERBOLIC FUNCTIONS



YEAR 12INDUCTION: PROBABILITY DISTRIBUTIONSIB HigherKnowledge QuizWhat is meant by equally likely outcomes?What is meant by independent events?Give an example of discrete data.What is meant by a “fair” dice?TREE DIAGRAMSA tree diagram can be used to show the outcomes of two (or more) events happening in succession.Example: A die is thrown three times. What is the probability that at least one of the rolls is a 6?QuestionsWhat if we wanted exactly one 6?What if we wanted exactly two 6s?What if we threw the die four times?What if we threw the die n times?Let’s explore this last question systematically. Find the probability that, in n rolls, we get:No 6sExactly one 6Exactly two 6sExactly three 6sExactly r lots of 6sSee the next page for an explanation of this task.ARRANGEMENTSIn the probability situation above, the order of the rolls is important. We often need to consider more than one route through a tree diagram to find all arrangements of outcomes that lead to the result we want. So for example, there are three ways of rolling a single 6 in three rolls: 6xx, x6x, xx6.In the Analysis course, you will study the theory of arranging items using binomial theorem in detail, but for now we can state the formula:We can obtain r objects from n objects in nCr waysnCr =n!r!n-r! , where n!=n×n-1×n-2×…×1While Analysis students will need to understand how to use this formula to compute arrangements, for Applications students it is enough to know that it is possible, because the GDC will be used for any calculations that involve arrangements like this. For the purpose of our investigation, we can simply represent the number of ways of getting r 6s from n rolls by with the notation nCr.THE BINOMIAL DISTRIBUTIONThe situation we have been exploring has a distinct set of characteristics that allow us to make a general probability formula. When we roll the die there are 2 possible outcomes: we get a 6, or we do not get a 6. We roll a set number of times, n, which means we can be specific about the number of 6s we get, with possible outcomes of 0 to n.Each roll is independent, so we can calculate the probability of each outcome through multiplication:P(rolling no 6s) = P(rolling n not 6s) = ( 56 )nP(rolling one 6) involves one 6 and (n-1) not 6s, i.e. ( 16 ) ( 56 )n-1. However as the one 6 could be positioned in any of the n rolls, the final probability would n ( 16 ) ( 56 )n-1P(rolling two 6s) involves two 6s and (n-2) not 6s, i.e. ( 16 )2 ( 56 )n-2.However as the two 6s could be positioned in any of the n rolls, we need to use binomial theorem to tell us how many possibilities there are for this. Arranging n items with 2 repeated items can be done in nC2 ways. So the final probability would be nC2 ( 16 )2 ( 56 )n-2We can begin to generalise this formula:P(rolling r 6s) involves r 6s and (n-r) not 6s, arranged in nCr possible ways,i.e. nCr ( 16 )r ( 56 )n-rIn Analysis, you can be required to find a binomial probability using this general formula, but it is NOT stated in the formula booklet. In both courses, you are required to use your GDC to calculate a binomial probability. In general a Binomial distribution is a suitable model when:there are a fixed number of trials, nthe trials are independent of each otherthere are only two outcomes at each trial (success and failure)the probability of success, p, is constant. The discrete random variable, X is “the number of successes”. If X has a Binomial distribution where the number of trials, n = 8 and the probability of success, p = 0.2, we write this as:X B(8, 0.2)QuestionsHow might a game where I do NOT want to score a 6 connect to the investigation you have carried out?How might these situations lead to a binomial model?A spinner with equal sized red, yellow and blue sections.A community with 55% of people right-handed.Some traffic lights which are red 48% of the time, yellow 4% of the time and green 48% of the time.When might the formula be more useful than your GDC?4114165323215Applications Example: In any group of students 30% wear glasses. A group of 5 students is chosen at random.(a) X = “number of students who wear glasses”We can model this by a Binomial distribution The parameters are p = 0.3 and n = 5so X ~ B(5, 0.3) (b)(i)P(X = 3)Using our GDC to calculate the binomial probability will be quicker than a tree diagram with 5 stages. (ii)P(X 2) Since the GDC will only calculate inclusive probabilities, i.e. P(X ≤ a), we first need to rewrite the problem. Since the distribution is discrete,P(X 2) = P(X ≤ 1)Analysis Example (non-calculator): In any group of students p% wear glasses. Several groups of 5 students are chosen at random.(a)Name and specify the probability distribution that would model the number of students who wear glasses within each group of 5.(b)Given that 1 in 32 of the groups have no students who wear glasses, find an estimate for the proportion of students who wear glasses in the population.Task: Write a flow map to explain the steps for using the GDC to find an exact binomial probability, and one for a cumulative binomial probability.Refer to your GDC manual.Task: Applications Task: A biased dice is rolled 20 times. Given that the dice will roll a 5 just 8% of the time, find:The probability that a quarter of the rolls are a 5.More than a quarter of the rolls are a 5.A game requires the player to get more than five 5s out of twenty rolls. If 900 people play the game, how many would you expect to win? X = “the number of 5s rolled”, X ~ B(20, 0.08)P(X = 5) = 0.0145 using the Bpd menu with x = 5P(X > 5) = P(X ≥ 6) = 0.00380 using the Bcd with lower 6 and upper 200.00380 x 900 = 3.42 or 3 peopleAnalysis Task: A biased dice is rolled 20 times. A game requires the player to roll at least one 5. Given that 10% of people will win the game, find the probability of getting a 5 on this biased dice.Hint: rather than working with at least one success, it will be easier to work with no failures! X = “number of 5s rolled”, X ~ B(20, p), where p represents the probability of getting a 5 in a single roll.Then the probability of winning isP(X ≥ 1) = 1 – P(X = 0) = 1 – (1-p)20Since 10% of people win the game, 1 – (1-p)20 = 0.1 0.9 = (1-p)20It is not reasonable that you would be asked to calculate this without a calculator! It works out as p = 0.00525. That is just a 0.5% chance of rolling a 5 on this biased dice. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download