 This video looks at some past paper questions on the binomial distribution. The first question I'm going to look at comes from the January 2008 paper, which is on the exciting topic of nuts and bolts. The question tells us that the probability of a bolt being faulty is 0.3 and that we're looking at a random sample of 20 bolts. The first part of the question asks us to calculate the probability that exactly two out of these 20 bolts are faulty. Well, the first thing is to create a random variable, X, which is equal to the number of faulty bolts. Because then we can say that X has the binomial distribution with 20 trials and 0.3 as the probability of success. It's 20 trials because there are 20 bolts and 0.3 because that's the probability of a bolt being faulty. Now the question is asking us for the probability that there are exactly two faulty bolts. And that's the probability that X is equal to 2. We know from the formula that that's equal to 20 choose 2 times 0.3 squared times 0.7 to the power of 18. Now 20 choose 2 is equal to 20 factorial divided by 2 factorial times 18 factorial, which can be simplified to 20 times 19 over 2 times 1. So the probability that X is equal to 2 is 20 times 19 over 2 times 1 times 0.3 squared times 0.7 to the power of 18, which is 0.0278 to four decimal places. The next part of the question asks for the probability of getting more than three faulty bolts. Well, we're still dealing with the same random variable X which has the binomial distribution with 20 trials and 0.3 as the probability of success. But this time it's asking us for the probability that X is greater than 3. Well, this is a probability that we'll want to work out using the tables, but we can't look up the probability that X is greater than 3 directly. Instead we need to realize that the chance of X being greater than 3 is 1 minus the probability that X is less than or equal to 3 because the opposite of being greater than 3 is being less than or equal to 3. We can find the probability that X is less than or equal to 3 by looking in the tables where n is equal to 20, p is equal to 0.3, and then following the row along from 3 to find the probability, which is 0.1071. So the chance of having more than three faulty bolts is one take away, 0.1071, which is 0.8929. The third part of the question introduces a new situation. Now we're told that bolts are sold in bags of 20 and that John buys 10 bags. The question is asking us for the probability that exactly six of these bags contain more than three faulty bolts. Now at this point we need to introduce a new random variable. We'll call it Y and make it equal to the number of bags with more than three faulty bolts. This has a completely different distribution. This time Y has the by-name of distribution with 10 trials and 0.8929 as the probability of success. 10 because there are 10 bags that John buys and 0.8929 because that was the answer to the previous question, the probability that an individual bag has more than three faulty bolts. Now the question is asking us for the probability that Y is equal to 6. And this time the formula tells us that that's 10 choose 6 times 0.8929 to the power of 6 times 0.1071 to the power of 4. 10 choose 6 is 10 factorial over 6 factorial times 4 factorial, which is 10 times 9 times 8 times 7 over 4 times 3 times 2 times 1. So now I've got a sum that we can do to calculate the probability that Y is equal to 6. And if you do it, you get the answer 0.0140 to 4 decimal places. And that's the answer to the question. The next question I want to look at is from the January 2002 paper and it's about organic food. It tells us that the probability of a diner in a restaurant choosing organic food is 40% and that we're looking at a sample of 20 diners. The first question asks for a suitable model to describe the number of diners who choose organic food. And then it wants to know the probability that that number is greater than 5 and less than 15. Well, the distribution of the number of people who choose organic food is binomial with 20 trials and 0.4 is the probability of success. 20 trials because there are 20 diners and 0.4 because that's the probability of somebody choosing to eat organic food. The question wants to know the probability that X is greater than 5 and less than 15. And this is probably something that we should think carefully about. For X to be greater than 5 and less than 15 means that it's 6, 7, 8 and so on up to 14. And note that we can get those numbers by taking all the numbers from 14 and below and then subtracting the ones from 5 and below. So the probability that X is greater than 5 and less than 15 is the probability that X is less than or equal to 14. Take away the probability that X is less than or equal to 5. We can look up both these probabilities in the tables. We need to find the table where n is 20, p is 0.4 and then follow on the rows from 14 which gives us 0.9984 and 5 which gives us 0.1256. So the probability that we're looking for is 0.9984 take away 0.1256 which is 0.8728. The last part of the question asks for the mean and the standard deviation of the number of diners who choose organic food. Well remember that X has the binomial distribution with 20 trials and 0.4 is the probability of success. So the mean will be 20 times 0.4 which is 8. The variance will be 20 times 0.4 times 0.6 which is 4.8. And therefore the standard deviation which is the square root of the variance will be the square root of 4.8 which is 2.19 to 3 significant figures. The last question I want to look at is from the June 2003 paper which is about biased dice. The first part of the question asks us to write down the conditions under which the binomial distribution will be a suitable model and you should know that there are 4 conditions so it's not surprising that this is for 4 marks. The first condition is that the number of trials is fixed. The second is that each trial should have the same two possible outcomes. The third is that the trials must be independent and the fourth is that the probability of success must be the same in each trial. Moving on to the next part of the question it tells us that when the die is thrown the number 5 is twice as likely to appear as any other number. Although the other faces are all equally likely to appear. We've got to find the probability. First of all that the first 5 will occur on the sixth throw. Okay well first of all we need to work out the probability of getting each number on the die. We can say that the probability of getting the numbers 1, 2, 3, 4 and 6 is p and that the probability of getting a 5 is 2p. But now we know that these probabilities must add up to 1 so we can say that 7p is equal to 1 and p is a seventh. That means that the probability of getting a 5, 2p is equal to 2 sevenths. Now the probability of getting the first 5 on the sixth throw is the probability of getting 5 non-fives followed by a 5. Now that's just going to be 5 sevenths to the power of 5 the probability of getting 5 non-fives in a row times 2 sevenths the probability that the last throw is a 5. And that probability is equal to 0.0531 to 3 significant figures. The last part of the question is asking us for the probability that out of the first 8th throws exactly 3 are fives. So for this let x be the number of fives it follows that x will have the binomial distribution with 8 trials and 2 sevenths as the probability of success 8 because there are 8 throws and 2 sevenths because that's the probability of getting a 5. We're being asked that x is equal to 3 and we know from the formula that that will be 8 choose 3 times 2 sevenths cubed times 5 sevenths to the power of 5 8 choose 3 is 8 factorial over 3 factorial times 5 factorial and that's 8 times 7 times 6 over 3 times 2 times 1 so that the sum that we have to do is 8 times 7 times 6 over 3 times 2 times 1 times 2 sevenths cubed times 5 sevenths to the power of 5 which is 0.2429 OK thank you for watching this video of exam questions It has to be said that exam questions on the binomial distribution often lead on to other topics and so these questions have not been particularly representative Nevertheless, I hope you found this useful revision