 This video looks at some past paper questions about the Poisson distribution. The first question is number five from the January 2009 paper. It's about a factory in which components are being made. Here's the question. It would be a good idea to stop the video at this point and have a go at the question yourself. Okay, so we're told that a factory produces components, one percent of which are defective, and that the components are packed in boxes of ten. Part A asks us to find the probability that the box contains exactly one defective component. Okay, well X will have the binomial distribution because it's the number of successes in a sequence of trials. The trials here being the components and success being that a component is faulty. In particular, X will have the binomial distribution with parameters ten and 0.01. Ten because that's the number of components, and 0.01 because that's the probability that a component is faulty. The question is asking us for the probability that X is equal to one, and we'll have to calculate this using the formula rather than the tables because the tables don't tell us anything about the case where the probability is 0.01. The formula says that this probability will be ten choose one times 0.01 to the power of one times 0.99 to the power of nine. But ten choose one will be just ten. Remember that n choose one is always just n. The first number along in Pascal's triangle after the one will be ten. And so the sum we have to do is ten times 0.01 to the power of one times 0.99 to the power of nine. And the answer to that is 0.0914 to four decimal places. The second part of the question asks us to find the probability that there are at least two defective components in the box. While we're looking at the same random variable which has the binomial distribution with parameters ten and 0.01, but this time we have to find the probability that X is greater than or equal to two. Okay, well being two or more is the opposite of being one or less. So the chance that X is greater than or equal to two will be one minus the probability that X is less than or equal to one. Or in other words one minus the sum of the chances that X is zero and X is one. Now the chance that X is zero will just be the probability of having ten working components in a row which is 0.99 to the power of ten. And that's 0.904382 and so on. The probability that X is one we just worked out and that's 0.091351 and so on. So the sum we have to do to find the answer is one take away the sum of these two numbers. And that gives us the answer 0.0043 to four decimal places. Okay, the last part of the question tells us to use a suitable approximation to find the probability that in a batch of 250 components between one and four are defective. So what we're going to do here is to use the Poisson distribution to approximate a binomial distribution. And the key thing we need to know is what's the expected number of defective components in this batch of 250? Well lambda is going to be 250 times 0.01. There are 250 components and the probability that any one of them is defective is 0.01. So you would expect there to be two and a half defective components. And Y will have the Poisson distribution with parameter 2.5. Now we're asked to find the probability that Y is greater than or equal to one and less than or equal to four. And that's the same as the probability that Y is less than or equal to four. Take away the probability that Y is zero. Because Y being one, two, three or four is the same as being less than or equal to four except for zero. At this point we can use the tables because the tables tell us about the case where lambda equals 2.5. So we find the column which is headed by lambda equals 2.5. We scan along from X equals four and from X equals zero to see the probability is 0.8912 and 0.0821. So the sum we have to do is 0.8912. Take away 0.0821 to get the final answer 0.8091 to four decimal places. And that's the final answer to this question. The next question I'd like to look at is number one from the January 2009 paper. And this is about daisies, which could look like this or like this. Here's the question. And again, it would be a good idea to pause the video at this point and to have a go at the question yourself. Okay, the question tells us that a botanist is studying the distribution of daisies in a field. And the field is divided into a number of equal sized squares. The mean number of daisies per square is assumed to be three. And the daisies are randomly distributed around the field. It's asking us to find the probability that in a randomly chosen square there'll be more than two daisies. Okay, so what we have here is a random variable with the Poisson distribution, because we're talking about a number of events in an interval of space. And it would be the Poisson distribution with parameter three, because the expected number of daisies in a square is three. The question is asking us to find the probability that x is greater than two, the probability that we get more than two daisies. And now x is greater than two means that x is three, four, five, and so on, which is the opposite of being not one or two. So the probability that x is greater than two will be one minus the probability that x is less than or equal to two. And we can look this up in the tables. We need to find the column headed by lambda equals three, and then scan along from where x is two to find the probability 0.4232. And then the sum that we need to do is one take away 0.4232, which gives us the answer 0.5768 to four decimal places. Next we have to work out the probability of getting either five or six daisies. So we're after the probability that x is equal to five or six. And this will be the probability that x is less than or equal to six. Take away the probability that x is less than or equal to four. Because five and six is all the numbers up to six, take away the numbers up to four. We can look up these two probabilities in the tables. Again, we're looking for the column where lambda equals three. And we scan along from where x is six and where x is four to find the probabilities 0.9665 and 0.8153. So the sum that we have to do is 0.9665, take away 0.8153. And that gives us the answer 0.1512 to four decimal places. Okay, the next part of the question tells us that the botanist has been to measure the number of daisies in each of 80 randomly selected squares. And it's giving us some statistics. We've got to work out the mean and the variance for the number of daisies per square. Okay, well the mean number of daisies will be sigma x divided by n. The total number of daisies in all the squares divided by the number of squares. And in this case that's 295 divided by 80, which is 3.69 to two decimal places. The formula for the variance is more complicated. This time we have to do sigma x squared divided by n, take away the square of sigma x over n. And if you put in the numbers, that means we have to do 1,386 divided by 80, take away the square of 295 over 80, which gives us the answer 3.73 to two decimal places. Okay, the question asks us to explain why this supports the choice of a Poisson distribution as a model for the number of daisies in a square. Now you'll probably remember that in a Poisson distribution both the mean and the variance are equal to lambda. So the mean and the variance are both the same with a Poisson distribution. In this case the two answers that we just got, 3.69 and 3.73 are very close together. So what we can say is that in a Poisson distribution the mean and the variance should be the same. And here the mean and the variance are very similar. So therefore it seems reasonable to use a Poisson distribution as a model. The final part of the question asks us to use our answer for the mean number of daisies to calculate the probability of getting exactly four daisies in a randomly chosen square. So the exact answer for the mean number of daisies was 3.6875. So what we have this time is a random variable that has the Poisson distribution with parameter 3.6875. We're asked to calculate the chance that x is equal to 4. And of course we can't use the tables now because the tables aren't going to tell us about the situation where lambda is 3.6875. Instead we'll have to use the formula which tells us that the probability is 3.6875 to the power of 4 times e to the power of minus 3.6875 divided by 4 factorial. And if you work that out you should get the answer 0.1929 to four decimal places. And that's the final answer to this question. Okay we're going to look at two more questions. The first one is question six from the January 2011 paper. And this one is about motorways. In particular it's about cars passing a toll booth on a motorway. You should probably pause the video at this point and have a go at this question. Okay so it's telling us that cars arrive at the motorway toll booth at an average rate of 150 per hour. And it's asking us to suggest a suitable distribution to the model the number of cars arriving at the toll booth in one minute. Well what we've got here is a Poisson distribution because you're looking at a number of events in an interval of time. The key thing is what's the parameter lambda? Well if 150 cars arrive every hour then the number of cars that you'd expect to arrive every minute is going to be 150 divided by 60. So lambda will be 150 divided by 60 which is 2.5. And therefore we'd expect x to have the Poisson distribution with parameter 2.5. We're asked to say what assumptions this makes. And there are four although there are only two marks here so presumably you only need to write down two of these. Firstly we must assume that cars arrive randomly. Secondly that they arrive at a constant average rate. Thirdly we must assume that cars arrive independently. And finally we must assume that they arrive one at a time. Okay now we have to use this model to find the probability that no cars arrive. So remember that we said that x has the Poisson distribution with parameter 2.5. And we're asked to find the probability that x is zero. Well we can put the numbers into the formula. The formula tells us that this is 2.5 to the power of zero times e to the power of minus 2.5 divided by nought factorial. And that can be simplified to just e to the power of minus 2.5. Because 2.5 to the power of zero is just one and nought factorial is one. So you've just got e to the power of minus 2.5. And if you work that out that's equal to nought point nought 821 to four decimal places. And secondly the question is asking us to find the probability that x is greater than three. Now being greater than three is being four five six seven and so on. And that's the opposite of being nought one two or three. So the probability that x is greater than three is one take away the probability that x is less than or equal to three. It's easiest to find this probability using the tables. So we find the column where lambda is equal to 2.5. And then we come across from where x is three to see the probability nought point seven five seven six. So the calculation we must do is one take away nought point seven five seven six. Which gives us the answer nought point two four two four to four decimal places. The last part of the question is a bit tricky. It's asking us to find a number m such that in a given four minute period the probability that x will be greater than m is equal to nought point nought four eight seven. Okay well first of all we'd better work out the distribution of x now because we've changed the time period from one minute to four minutes. Now in one minute we expected to get two and a half cars. So in a four minute period you'd expect to get four lots of two and a half cars. In other words ten cars. So in this case x will have the Poisson distribution with parameter ten. Now let's stop and think what we have to do next. We have to find a value of m such that the probability that x is greater than m is equal to nought point nought four eight seven. Now in an ideal world nought point nought four eight seven would be a probability that we would find in a probability table. And all we would have to do is to see the value of m that would give us this probability. But the probability tables never tell us about the chance that x is greater than something. They always tell us about the probability that x is less than or equal to something. So after all nought point nought four eight seven won't be something that we'll find in a probability table. Instead we need to think about the probability that x is greater than m and express it in terms of the probability that x is less than or equal to something. So let's think about that. Now x being greater than m means that x is m plus one or m plus two or m plus three and so on. That's the opposite of being m or m minus one or m minus two and so on. More simply the event that x is greater than m is the opposite of the event that x is m or less than m. So the probability that x is greater than m will be one take away the probability that x is less than or equal to m. So what we can say is that one take away the probability that x is less than or equal to m will be nought point nought four eight seven. And if you think about it, this means that the probability that x is less than or equal to m will be 1 minus 0.0487, which is 0.9513. You can see that we need to subtract something like 0.9513 from 1 in order to get 0.0487. Okay, well now we've got something where we can use the tables. Because if we look up the right value for m in the tables, we would get the answer 0.9513. So all we have to do is to work backwards. We find the probability 0.9513. And then we see what value of m is needed in order to get this answer. So let's look at the tables. We find the column where lambda is equal to 10. And if you look down that column, you'll see the probability 0.9513. If you scan along the row, you'll see that it begins with the number 15. And what this shows us is that m has to equal 15. 15 is the number that you need to look up in order to get the answer 0.9513. The last question I want to look at is question 8 from June 2009. And this is about the manufacture of cloth. Here's the question. As before, you should pause the video at this point and think about how you would answer this question. The question tells us that a cloth manufacturer finds faults in the cloth at the rate of 2 every 15 meters. And we're asked to find the probability of getting exactly four faults in a 15-meter length of cloth. OK, well, the number of faults in a 15-meter length of cloth will have the Poisson distribution, because here we're talking about a number of events in an interval of space. And we know that the expected number of faults is 2. So x will have the Poisson distribution with parameter 2. The question is asking us to find the probability that x is equal to 4. It's easiest to use the formula to work this out. The formula tells us that this is 2 to the power of 4 times e to the power of minus 2 divided by 4 factorial. And if you tap those numbers into your calculator, you'll get the answer 0.0902 to four decimal places. Next, we have to find the probability of getting more than 10 faults in 60 meters of cloth. OK, well, the number of faults in 60 meters of cloth will have a different distribution. This time, the expected number of faults will be more than 2. 60 meters is 4 lots of 15 meters. So the expected number of faults will be 4 lots of 2. In other words, you would expect to see 8 faults in 60 meters of cloth. So this time, the number of faults will have the Poisson distribution with parameter 8. The question is asking us to find the probability that x is greater than 10. And remember that being greater than 10 is being 11, 12, 13, and so on. And that's the opposite of being 10 or less. So the probability is one take away, the probability that x is less than or equal to 10. We can look that up in the tables. We need to find the column headed by lambda equals 8 and then look along from 10 so that we find the probability 0.8159. So the calculation we must do is one take away 0.8159, which is 0.1841 to four decimal places. The next part of the question is a tricky one. The manufacturer is going to choose a length of cloth in order to get a certain probability of having no faults. Now, in this situation, we need to know the distribution for the number of faults in the cloth. Now, the expected number of faults in one meter of cloth is going to be 2 divided by 15. There are two faults in 15 meters on average, so the number of faults in one meter of cloth is going to be 2 divided by 15. And that's 2.15. So if you expect 2.15 of a fault in one meter, then the number of faults you'd expect in x meters would be 2.15 times x. So now we can say that the number of faults in this length of cloth will have the Poisson distribution with parameter 2.15 of x. The number of faults in a piece of cloth of length x meters will have the Poisson distribution where the expected number of faults is 2.15 of x. Now let's stop and think again. We need to find a value of x such that the probability of getting no faults is equal to 0.80. And the obvious way of doing this is to find an expression for the probability of getting no faults in terms of x so that we can make an equation. We'll get an expression for the probability of getting no faults. We'll set this equal to 0.8. And then we'll solve it for x. We're not going to be able to use the tables in this situation, by the way. Because if you look at the answer, you can see that x is equal to 1.7. And therefore lambda is a horrible number which won't appear in the tables. OK, so let's find an expression for the probability that x is equal to 0. The chance that x is equal to 0 is always lambda to the power of 0 times e to the power of minus lambda over 0 factorial. But since lambda to the power of 0 is equal to 1 and 0 factorial is equal to 1, this can be simplified to e to the power of minus lambda. Now in this case, lambda is equal to 2.15 of x. So what we can say is that the probability that x is equal to 0 is e to the power of minus 2.15 of x. And now we can make an equation. We can say that e to the power of minus 2.15 of x is equal to 0.80. And all that's left to do is to solve this equation. Well, we can solve it by taking logs of both sides. Taking logs tells us that minus 2.15 of x is equal to log 0.8, the natural logarithm of 0.8. And then all we have to do is to multiply both sides by minus 15 over 2. And it tells us that x is equal to minus 15 over 2 times the natural logarithm of 0.80. And if you type those numbers into your calculator, you get 1.7 to two significant figures, which is the answer that the question gave us. OK, there's one more part of the question. It tells us that the retailer sells 1,200 pieces of the cloth. He makes a profit of 60p on anything that doesn't have a fault and a loss of 1 pound 50 on anything that does. So we've got to find the expected profit. Well, first of all, the expected number of pieces of cloth with no faults will be 1,200 times 0.8. There are 1,200 pieces, and the probability of having no faults is 0.8. So that gives us an answer of 960. We expect 960 pieces with no faults. Similarly, the expected number with faults is going to be 1,200 times 0.2, which is 240. So now we can say that the expected profit will be 960 times 60p. Take away 240 times 1 pound 50. On average, he's going to make a profit of 60p on 960 pieces and a loss of 1 pound 50 on 240 pieces. The answer to that is 216 pounds, which is the final answer to this question. OK, that's the final question. I hope you found this video useful. Thank you very much for watching.