The Poissondistribution ✏ ✒ ✑ 5.3 Introduction In this block we introduce a probability model which can be used when the outcome of an experiment is a random variable taking on positive integer values and where the only information available is a measurement of its average value. This has widespread applications in analysing traffic flow, in fault prediction on electric cables, in the prediction of randomly occurring accidents etc. ✛ ✘ Prerequisites Before starting this Block you should . . . ① understand the concepts of probability and probability distributions. ✚ Learning Outcomes ✙ Learning Style After completing this Block you should be able To achieve what is expected of you . . . to . . . ✓ recognise and use the formula for probabilities calculated from the Poisson model ☞ allocate sufficient study time ✓ use the recurrence relation to generate a succession of probabilities ☞ briefly revise the prerequisite material ✓ use the Poisson model to obtain approximate values for binomial probabilities ☞ attempt every guided exercise and most of the other exercises We need not calculate each probability directly. The Poisson distribution Suppose that it has been observed that. 180 cars per hour pass a specified point on a particular road in the morning ‘rush hour’. If the occurrences of an event are relatively rare.1. according to the Poisson model. Key Point The Poisson Probabilities If X is the random variable ‘number of occurrences in a given interval’ for which the average rate of occurrences is λ then. for ease of calculation we use the recurrence relationship P (X = 1) = P (X = r) = λ P (X = r − 1) for r ≥ 1. the probability of r occurrences in that interval is given by P (X = r) = e−λ λr r! Now do this exercise Using the Poisson distribution write down the formulae for P (X = 0). What is the probability of this latter event occurring? We cannot use the binomial model here since there is no fixed number of cars which will pass the point and we have no estimate of p. on average. noting that 0! = 1. Answer We have calculated P (X = 0) to P (X = 5) when λ = 2 and presented the results to 4d.2707 0.the Poisson distribution .1353 0. P (X = 2). Under these assumptions the Poisson model can be developed. r Engineering Mathematics: Open Learning Unit Level 1 5.0902 0.0361 Table 1 Notice how the values for P (X = r) increase and then decrease relatively rapidly (due to the decreasing exponential term e−λ ). P (X = 6). P (X = 2) = P (X = 1) P (X = 3) = P (X = 2) . We must use a different probability model .1804 0. r 0 1 2 3 4 5 P (X = r) 0. we may assume that the number of occurrences in any small interval (of time or space as appropriate) is independent of the number of occurrences in any other small interval. (We omit the derivation).2707 0.p. We can use the following relations (which can be checked from the formulae for P (X = r)): λ λ λ P (X = 0) .3: Applied Probability Distribution 2 . in Table 1. 1 2 3 In general. Due to impending road works it is estimated that congestion will occur closer to the city centre if more than 5 cars pass the point in any one minute. In this example two of the probabilities are equal and this will always be the case when λ is an integer. etc. P (X = 1).to handle situations where the only information is an average rate. starting from P (X = 0). P (more than 5 cars in one minute) = 1 − P (5 cars or less arrive in one minute) Thus P (X > 5) = 1 − P (X ≤ 5) = 1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3) − P (X = 4) − P (X = 5) Each probability can be calculated by the recurrence relation above. Remember that if X is a random variable following a binomial distribution then E(X) = np and V (X) = npq. We shall see later that the Poisson model can then be used as a convenient way of approximating the probabilities arising in the binomial model. When the number of trials n in the binomial model is increased without bound and p (the probability of a success on an individual trial) tends to zero in such a way that np remains at a value (= λ) say then the binomial model becomes more like the Poisson model introduced here. Then we calculate P (X = 0) to P (X = 5) and finally the required probability. The decision is either to use an accurate model with tedious (and probably inaccurate) calculations or to use an approximate model with more easily-calculated probabilities which will be approximations only. n → ∞) of the binomial distribution and putting λ = np. 2. let X be the random variable ‘number of cars arriving in any minute’ . We cannot give a hard-and-fast rule as to how 3 Engineering Mathematics: Open Learning Unit Level 1 5. Calculate the individual probabilities to 5 d. This is best done by noting that. Calculate the probability that more than 5 cars arrive in one minute. First we convert the information on the average rate (cars per hour) to obtain a value for λ (cars per minute). If p is quite small then q 1 and npq np. the calculation of the (binomial) probabilities can be tedious and inaccurate.3: Applied Probability Distribution . Poisson approximation to the binomial distribution We have already noted that the Poisson model is obtained by taking the special limit (p → 0. referred to at the beginning of this section. the standard deviation is λ. we find: E(X) = λ and V (X) = λ √ Clearly.p. then. because of its relation to the binomial distribution (in which E(X) = np = λ and V (X) = npq = λ(1 − p). Answer Relation to the binomial distribution It can be shown that the Poisson distribution may be obtained from the binomial distribution by taking a particular limit. Since V (X) E(X) it is possible to use a Poisson model to generate probabilities approximately equal to their binomial counterparts.Now do this exercise Returning to the car example. Note that if X is a random variable whose probabilities are given by the Poisson model then. In some problems for which the binomial model is appropriate. and the final number to 4 d.p. It is believed that 1 item in 2000 on average is substandard.p.. Answer Now do this exercise Now choose a suitable value for λ in order to use a Poisson model to approximate the probabilities. suppose mass-produced items are packed in boxes of 1000. Answer We have obtained the same answer to 4 d. Hint: don’t evaluate the final probability until you have a compact expression for it. What is the probability that a box contains more than 2 defectives? 1 The correct model is the binomial distribution with n = 1000. as the exact binomial calculation. Engineering Mathematics: Open Learning Unit Level 1 5. we can simply say ‘the smaller the better’ so long as np remains of a sensible size. We shall not always be so lucky. Answer Now do this exercise Now recalculate the probability that there are more than 2 defectives using the Poisson distribution with λ = 12 .3: Applied Probability Distribution 4 . As an example. 2000 Now do this exercise Using the binomial distribution write down P (X = 0).small p should be for the Poisson approximation to be reasonable. p = 2000 (and q = 1999 ). essentially because p was so small. P (X = 1) and calculate P (more than 2 defectives). every item in the batch is inspected. and the manufacturer normally produces 2% of defective articles. If 250 litres of water are known to be polluted with 106 bacteria what is the probability that a sample of 1cc of the water contains no bacteria? 3. (b) less than 5? If. If inspection costs are 75p per hundred articles. By agreement with a customer the following method of inspection is adopted: A sample of 100 items is drawn at random from each batch and inspected. Large sheets of metal have faults in random positions but on average have 1 fault per 10m2 . If more than 4 defectives are found. (b) What is the largest number. find the average inspection costs per batch. Suppose vehicles arrive at a signalised road intersection at an average rate of 360 per hour and the cycle of the traffic light is set at 40 seconds. What is the probability that a sheet 5m×8m will have at most one fault. of transistors that can be put in a box so that the probability of no defectives is at least 1/2. A manufacturer sells a certain article in batches of 5000. N . Answer 5 Engineering Mathematics: Open Learning Unit Level 1 5. Previous results indicate that 1 in 1000 transistors are defective on average (a) Find the probability that there are 4 defective transistors in a batch of 2000.More exercises for you to try 1. In what percentage of cycles will the number of vehicles arriving be (a) exactly 5.3: Applied Probability Distribution . then the batch is accepted by the customer. after the lights change to green. 5. 2. there is time to clear only 5 vehicles before the signal changes to red again. what is the probability that waiting vehicles are not cleared in one cycle? 4. If the sample contains 4 or fewer defective items. 3: Applied Probability Distribution 6 .End of Block 5.3 Engineering Mathematics: Open Learning Unit Level 1 5. 3: Applied Probability Distribution .λ0 1 λ P (X = 1) = e−λ × = λe−λ = e−λ × ≡ e−λ 0! 1 1! 2 2 6 6 λ λ λ λ = e−λ = e−λ P (X = 2) = e−λ × P (X = 6) = e−λ × 2! 2 6! 720 P (X = 0) = e−λ × Back to the theory 7 Engineering Mathematics: Open Learning Unit Level 1 5. 22404 0.22404 0.149361 0.p).91608 Then P (more than 5) = 1 − 0. We present the results in a table.λ = 3 cars/minute. First.168031 0.0839 (4 d. P (X = 0) = e−3 r 0 1 2 3 4 5 Sum P (X = r) 0.04979 0.3: Applied Probability Distribution 8 .08392 = 0.91608 = 0.10082 0. Back to the theory Engineering Mathematics: Open Learning Unit Level 1 5. 0902.3: Applied Probability Distribution .) 2000 Hence P (more than 2 defectives) 1 − 0. Back to the theory 9 Engineering Mathematics: Open Learning Unit Level 1 5.p.9098 = 0.9098 (4d. 1000 1999 P (X = 0) = 2000 999 999 1999 1 1 1999 P (X = 1) = 1000 × = 2000 2000 2 2000 ∴ P (X = 0) + P (X = 1) = 1999 2000 999 1999 1 + 2000 2 = 1999 2000 999 × 2999 0. λ = np = 1000 × 1 1 = 2000 2 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 5.3: Applied Probability Distribution 10 . p. P (X = 0) = e− 2 .) 2 Hence P (more than 2 defectives) 1 − 0. 1 Back to the theory 11 Engineering Mathematics: Open Learning Unit Level 1 5.9098 (4 d.0902.1 1 P (X = 1) = e− 2 2 3 −1 ∴ P (X = 0) + P (X = 1) = e 2 = 0.3: Applied Probability Distribution .9098 = 0. In 40 seconds we expect 4 vehicles ∴ λ=4 (a) P (exactly 5) = λ5 e−λ /5! = 0.5 ∴ N = 693. 1 =2 1000 P (X = 4) = λ4 e−λ /4! = 16e−2 /24 = 0. P (greater than 5) = 1 − P (exactly 5) −P (less than 5) = 1 − 0. (a) Poisson approximation to Binomial λ = np = 2000.6288 = 0.0183 3.2148 4.0916 2. 12 .6288 3 3 Vehicles will not be cleared if more than 5 are waiting.6% of cycles λ 2 λ3 λ4 −λ 4 + + 1+λ + (b) P (less than 5) = e 2! 3! 4! 32 32 = e−4 1 + 4 + 8 + + = 0. Poisson Process.09022 (b) λ = N p = N/1000.3: Applied Probability Distribution ∴ −N = ln(0. P (X = 0) = λ0 e−λ = e−λ = e−N/1000 0! e−N/1000 = 0. In a sheet size 40m2 we expect 4 facults ∴ P (X = r) = λr e−λ /r! λ=4 P (X ≤ 1) = P (X = 0) + P (X = 1) = e−4 + 4e−4 = 0.15629 − 0.e.147 Engineering Mathematics: Open Learning Unit Level 1 5.15629 i. In 1cc we expect 4 bacteria(= 106 /250000) ∴ λ=4 P (X = 0) = e−4 = 0. in 15.1.5) 1000 choose N = 693 or less. 02) = 2 P (4 or fewer defectives in sample of 100) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = e−2 + 2e−2 + Inspection costs 22 −2 23 −2 24 −2 e + e + e = 0. P (defective) = 0.0526 E(Cost) = 75(0.5p Back to the theory 13 Engineering Mathematics: Open Learning Unit Level 1 5. Poisson approx.5. to Binomial λ = np = 100(0.02.3: Applied Probability Distribution .0526) = 268.947347 0.947347) + 75 × 50(0.947347 2 3! 4! Cost c 75 75 × 50 P (X = c) 0.