STEP Support Programme. Statistics STEP Questions

STEP Support Programme Statistics STEP Questions This is a selection of STEP I and STEP II questions. The specification is the same for both papers, with STEP II questions designed to be more difficult. 2010 S1 Q12 1 Preparation The expectation of a discrete random variable is defined by: E(X) = n P (X = n) where the sum is taken over all possible values of n. The discrete random variable X is defined by: { kx for x = 1, 2, 3, 4, P (X = x) = 0 otherwise. (a) Find the value of k. (b) Find P (X 3). (c) Find E(X). I roll a normal six-sided die 1 until I roll a 6; then I stop. (a) Find the probability that I roll a 6 on the third roll. (b) Find that probability that I need more than five rolls to get a 6. You could do 1 ( P(X = 1) + P (X = 2) + + P (X = 5) ) but it is easier to think about what must have happened on the first five rolls if you need more than five rolls to get a 6. (iii) Find the value of the infinite sum 2 + 0.2 + 0.02 +. (iv) Find the maximum value of y = x(4 x). 1 Apparently, the use of die as the singular form of dice is normal in America and in maths books, and certainly on STEP papers. But, at least in conversation, the British never say die. Statistics STEP Questions 1

2 The Probability question A discrete random variable X takes only positive integer values. Define E(X) for this case, and show that E(X) = P (X n). n=1 I am collecting toy penguins from cereal boxes. Each box contains either one daddy penguin or one mummy penguin. The probability that a given box contains a daddy penguin is p and the probability that a given box contains a mummy penguin is q, where p 0, q 0 and p + q = 1. Let X be the number of boxes that I need to open to get at least one of each kind of penguin. Show that P(X 4) = p 3 + q 3, and that Hence show that E(X) 3. E(X) = 1 pq 1. Discussion Even if you cannot do the first part of the question, you can still do the rest. It may be helpful to note that, for example, 3 P (X = 3) = P (X = 3) + P (X = 3) + P (X = 3). For P (X 4) you could consider 1 P (X 3) but this is probably not the easiest way. For E(X) be careful with the n = 1 case, it does not fit the same pattern as the other values of n. Statistics STEP Questions 2

2009 S1 Q13 3 Preparation This question involves arrangements, and you may like to look back to Assignment 6 before you start. You can leave big numbers in terms of factorials. 6 women (A, B, C, D, E and F) stand in a line. In how many ways can they arrange themselves in this line? There are 6 different positions for A, then 5 for B etc. The 6 women are joined by 4 men (G, H, I and J). In how many ways can the 10 people arrange themselves? (iii) The 4 men wish to stand together. We can imagine them being together inside a rope ring. (a) In how many ways can the 4 men arrange themselves in a line inside their rope? (b) Imagine now that the rope is one Person so that there are 7 people i.e. the 6 women and the 1 roped-off collection of men. How many ways can we arrange these 7 people? (c) How many ways are there of arranging the 10 people so that the men stand together? (d) Show that the probability that all 4 men stand together is 1/30. (iv) What is the probability that all the women (but not necessarily the men) stand together? (v) What is the probability that all the men stand together and all the women stand together? Note that this is not the previous two answers multiplied together as the events are not independent. Statistics STEP Questions 3

(vi) Find the probability that no men stand together by following these steps. (a) First arrange the 6 women. How many ways can we do this? (b) There are now 7 gaps between the women which we can slot the first man into, like this: W 1 W 2 W 3 W 4 W 5 W 6 where the arrows represent the gaps. Once a man fills up a gap then it is no longer a gap and there are only 6 gaps for the second man. Find the number of ways you can arrange the 4 men in the 7 gaps. (c) Find hence find the probability that no two men are together. (vii) Find the probability that there will be a woman at each end of the line, by following these steps. (a) How many different ways can we choose the two women to be at the ends? (Consider how many choices you have for the first end and then how many for the second end). (b) How many ways can you arrange the 8 people left? (c) Use your two previous answers to find the probability that there is a woman at each end of the line. 4 The Probability question I seat n boys and 3 girls in a line at random, so that each order of the n + 3 children is as likely to occur as any other. Let K be the maximum number of consecutive girls in the line so, for example, K = 1 if there is at least one boy between each pair of girls. Find P(K = 3). Show that P(K = 1) = n(n 1) (n + 2)(n + 3). (iii) Find E(K). Discussion You might like to simplify your final answer, though the question doesn t ask you to do so. More overleaf... Statistics STEP Questions 4

If you found the last question interesting, you might like to try this STEP question as well: 5 1995 S1 Q12 A school has n pupils, of whom r play hockey, where n r 2. All n pupils are arranged in a row at random. What is the probability that there is a hockey player at each end of the row? What is the probability that all the hockey players are standing together? (iii) By considering the gaps between the non-hockey-players, find the probability that no two hockey players are standing together, distinguishing between cases when the probability is zero and when it is non-zero. Discussion The last phrase distinguishing between cases when the probability is zero and when it is non-zero is a rather poorly phrased attempt to point out that for some values of n and r the probability is zero, i.e. it is not possible for all the hockey players to be separated. A little bit of thinking should give a relationship between n and r for when the hockey players cannot possibly be separated. The number of ways of choosing r objects from a total of n different objects is written as n C r or ( n r). You may find this notation useful. n!, which is r!(n r)! Statistics STEP Questions 5

2010 S1 Q13 6 Preparation By first writing 9 x = (3 2 ) x, solve 3 2x 3 x 2 = 0, giving your answer(s) in terms of logarithms. Solve the inequality 1 1 x > x. Squaring both sides can be a valid method but only in certain circumstances such as when you know both sides are positive. The Poisson distribution measures the number of occurrences of an event in a given time interval. It was first used by Ladislaus Josephovich Bortkiewicz to model the number of deaths of Prussian cavalry-men by horse kicks in a year. A Poisson random variable satisfies the following conditions: I Occurrences are independent. II The probability of an occurrence during any time interval is proportional to the duration of the time interval. As well as modelling the number of occurrences in a given time interval it can be used to model the number of occurrences in a given space interval. Some applications are the number of car accidents in a mile of road, the number of people joining a queue every 5 minutes and the number of hairs in a burger. The number of occurrences in a given time interval is given by: P(X = n) = e λ λ n where n is an integer, with n 0, λ is the mean number of occurrences in the given interval and (by convention) 0! = 1. Note that the sum of all the probabilities is given by: n! e λ λ n n=0 n! = e λ n=0 λ n n! = e λ e λ = 1. For the last equality, we used the exponential series. You may like to show that E(X) = λ. (iii) The number of patients arriving at a drop-in chocoholics centre follows a Poisson distribution with an average rate of 5 an hour (so λ = 5). Don t bother to evaluate the exponentials or powers or factorials in the following questions. (a) Find the probability that exactly 3 patients arrive between 9am and 10am. (b) Find the probability that at least 1 patient arrives between 10am and 11am. Statistics STEP Questions 6

(c) The centre closes for lunch between 1pm and 2pm. It then runs for 3 more hours in the afternoon. Find the probability that no-one turns up for the first hour after lunch, then exactly 2 people turn up between 3pm and 4pm and then exactly 6 people turn up in the hour before closing. Disjoint time intervals are independent, so you can multiply the probabilities. (iv) If the number of occurrences in a time interval can be modelled as a Poisson distribution with a mean of λ, then the number of occurrences in a time interval twice as long can be modelled as a Poisson distribution with mean 2λ, and so on. The number of calls to Kalculus Kids is Poisson distributed with, on average 8 calls an hour. Find the probabilities of: (a) exactly 5 calls in one hour. (b) exactly 10 calls in two hours. (c) exactly 2 calls in 30 mins. (d) Fewer than 2 calls in 15 mins. (v) If you have two independent Poisson random variables, X with mean λ and Y with mean µ, then the random variable X + Y also has a Poisson distribution, with mean λ + µ. A company making gizmos accepts orders on-line or by phone. Both of these follow a Poisson distribution: telephone orders with a mean of 2 per day; and on-line orders with a mean of 5 per day. (a) What is the probability that they receive 10 orders on one day? (b) What is the probability that they receive fewer than 3 orders in one day? Statistics STEP Questions 7

7 The Probability question The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean λ texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is p, show that pe 2λ e λ + 1 = 0. Given that 4p < 1, show that there are two positive values of λ that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means λ 1 and λ 2 texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also p, find an expression for λ 1 + λ 2 in terms of p. Find the probability, in terms of p, that she waits between 1 and 2 hours in the morning to receive her first text. Discussion Note that λ > 0 if and only if e λ > 1. Statistics STEP Questions 8

2010 S2 Q12 8 Preparation You are probably familiar with discrete random variables, and may have met the Normal Distribution which is a special continuous random variable. A continuous random variable has a probability density function which can be used to find the probability that the random variable s value falls in a particular range. The probability that the random function X lies in the range a X b is given by: b where f(x) is the probability density function. a f(x)dx Since the total probability has to be 1, the total area under the graph y = f(x) must be 1. You will sometimes have to use this fact, for example to find the value of a constant in f(x). The expectation of a continuous random variable is given by: E(X) = x f(x)dx. The Median of a continuous random variable is the value M such that: M f(x)dx = M f(x)dx = 1 2, i.e. M splits the graph up so that the area each side is 1 2. Remember that f(x) might be equal to zero for some ranges of x. A continuous random variable X has probability density function f(x), where (a) By using 2 0 f(x) = f(x)dx = 1, find k. { kx for 0 x 2 0 otherwise. You don t have to actually integrate, you could use the formula for the area of a triangle instead. (b) Find P(1 X 2). (c) Find E(X). The integral will be from x = 0 to x = 2 as the probability density function is zero elsewhere. (d) Find the median, M, of X. Statistics STEP Questions 9

A continuous random variable X has probability density function f(x), where (a) Verify that 5 3 0 f(x) dx = 1. x 2 for 0 x 1 f(x) = 1 for 1 x 5 3 0 otherwise. You will need to do two separate integrals, one for 1 2 x 1 and one for 1 x 5 3. (b) Find P ( 1 2 X 4 3). Again you will need two integrals. (c) Find E(X). Again you will need two integrals. (d) Find the median, M. It would probably be a good idea to find median lies. 1 0 f(x)dx first to see which side of x = 1 the 9 The Probability question The continuous random variable X has probability density function f(x), where a for 0 x < k f(x) = b for k x 1 0 otherwise, where a > b > 0 and 0 < k < 1. Show that a > 1 and b < 1. Show that E(X) = 1 2b + ab 2(a b). Show that the median, M, of X is given by M = 1 2a expression for the median if a + b 2ab. if a + b 2ab and obtain an (iii) Show that M < E(X). Discussion For the very first part a well drawn sketch, with a few words of explanation, should be enough to show the required result. You will need to find k in terms of a and b. For part the answers depend on which side of x = k the median lies. When trying to show that M < E(X), it will probably be easier to show that M E(X) < 0. Statistics STEP Questions 10

1999 S2 Q12 10 Preparation Sketch the graph y = 3x 5 2x. Find the maximum and minimum values of y if x lies in the interval 1 x 2. To find the horizontal asymptote it may be helpful to divide both the numerator and denominator by x. Sketch the graph y = 1 x 5 2x. Find the maximum and minimum values of y if x lies in the interval 1 x 2. (iii) The diagram below is called a Venn diagram. In a Venn diagram, probabilities are represented by areas, and the blue rectangle (representing the whole sample space) has area 1. The area in the red circle represents P(A) and the area in the green circle. represents P(B). The probability of A and B both happening, which is written as P(A B), is represented by the area which lies in both the red and the green circles (the bit with a yellow star in it). The probability that either A or B, or both, occur is written as P(A B) and is represented by the whole area covered by the red and green circles. If A and B were mutually exclusive then the two circles do not overlap and P(A B) = P(A) + P(B). If A and B are not mutually exclusive, then we have to ensure that we do not count the intersection twice, so: P(A B) = P(A) + P(B) P(A B). Statistics STEP Questions 11

The same diagram can be used to find conditional probabilities. In order to find the conditional probability that A happens given that B happens, we look at the fraction of the area of the B circle that is also covered by the A circle. This gives the conditional probability of A happening given that B happens as: P(A B) = P(A B) P(B). In order to calculate conditional probabilities when events A and B are not independent, we use: P(A B) P(A B) P(A B) = and P(B A) = P(B) P(A) to give: and hence: P(A B)P(B) = P(B A)P(A) P(A B) = P(B A)P(A) P(B) This last result is known as Bayes Theorem. (a) Given that P(A) = 0.3, P(B) = 0.5, that the events A and B are independent, find P(A B). Find also P(A B) and P(B A). If the events are independent then P(A B) = P(A) P(B). (b) Events A and B (not necessarily independent) satisfy: Find: P(A B) and P(A B). P(A) = 0.6, P(B) = 0.5, P(B A) = 0.4. (c) 1% of Martians at age forty who participate in routine screening for tentacle rot actually have tentacle rot. It is known that 80% of Martians with tentacle rot will get a positive result in the screening. 10% of Martians without tentacle rot will also get a positive result. A Martian of age forty had a positive result in a routine screening. What is the probability that she actually has tentacle rot? You can do this is various ways. You could just consider a population of 1000 Martians. You could draw a Venn diagram. You could use Bayes theorem, in which case you might find the result P(B) = P(B A)P(A) + P(B A )P(A ) useful. (Here A is the complement of A, i.e. the event A does not happen. We suggest that you try all three ways and see which you like best. Statistics STEP Questions 12

11 The Probability question It is known that there are three manufacturers A, B, C, who can produce micro chip MB666. The probability that a randomly selected MB666 is produced by A is 2p, and the corresponding probabilities for B and C are p and 1 3p, respectively, where 0 p 1 3. It is also known that 70% of MB666 micro chips from A are sound and that the corresponding percentages for B and C are 80% and 90%, respectively. Find in terms of p, the conditional probability, P(A S), that if a randomly selected MB666 chip is found to be sound then it came from A, and also the conditional probability, P(C S), that if it is sound then it came from C. A quality inspector took a random sample of one MB666 micro chip and found it to be sound. She then traced its place of manufacture to be A, and so estimated p by calculating the value of p that corresponds to the greatest value of P(A S). A second quality inspector also a took random sample of one MB666 chip and found it to be sound. Later he traced its place of manufacture to be C and so estimated p by applying the procedure of his colleague to P(C S). Determine the values of the two estimates and comment briefly on the results obtained. Discussion A lot of reading. But you shouldn t be put off by questions that look long on the page: often (and especially for mechanics and probability), they are long because there are quite a few things to say that take up space, but hardly need saying (stuff about no wind resistance, randomness, etc); and sometimes it is because some new concept is being explained once you have grasped the concept, the question becomes fairly straightforward. Statistics STEP Questions 13

2006 S2 Q13 12 Preparation A group of 5 children, aged 2, 4, 6, 8 and 10, line up randomly (so that each order in the line is equally likely) for birthday cake. (a) Find the probability that the youngest child is first in the queue. (b) Find the probability that the second child in the queue is the oldest given that the first child is the youngest. I create a four-digit number by placing the digits 1, 2, 3 and 4 in random order (each order being equally likely). (a) How many such numbers can I create? (b) How many of these numbers are divisible by 4? There are various ways of doing this the low-tech approach is to write out all the different options. (c) What is the probability that I create a number bigger than 2413? (d) If the first digit of my number is 1, what is the probability that the number is divisible by 4? 13 I know that ice-creams come in n different sizes, but I don t know what the sizes are. I am offered one of each in succession, in random order. I am certainly going to choose one - the bigger the better - but I am not allowed more than one. My strategy is to reject the first ice-cream I am offered and choose the first one thereafter that is bigger than the first one I was offered; if the first ice-cream offered is in fact the biggest one, then I have to put up with the last one, however small. Let P n (k) be the probability that I choose the kth biggest ice-cream, where k = 1 is the biggest and k = n is the smallest. Show that P 4 (1) = 11 24 and find P 4(2), P 4 (3) and P 4 (4). Find an expression for P n (1). Discussion The expression in part will be in the form of a sum. Statistics STEP Questions 14

2008 S2 Q13 14 Preparation I have a row of 5 current buns, one of which has a cherry on top. If I close my eyes and pick two at random, what is the probability that I pick the bun with a cherry on top? Now I have n current buns, of which one has a cherry on top. If I pick r buns, what is the probability that I pick the one with a cherry on top? You are asked to do the same thing below in part (iv), but here you should use a simple argument. (iii) The number of ways of choosing r objects from a total of n different objects is n! r!(n r)!, which is written as n C r or ( n r). These are also binomial coefficients i.e. the coefficient of x r in the expansion of (1 + x) n. A team of 5 has to be chosen from a class of 10 swimmers, of whom exactly 2 are male. Find the probability that a team of 5 chosen at random (the probability of each choice being equally likely) will include the 2 males. You will need to find the total number of ways of choosing 5 from the class of 10, and also (to work out the number of these in which there are two males) the number of ways of picking 3 from the remaining 8 once you have picked the 2 males. (iv) Use the n C r notation to find the probability that, if I pick r current buns from a total of n, I get the one with a cherry on top. You need to consider the total number of ways of picking r buns out of the n as well as the number of ways that include the 1 with a cherry on top. (v) I have n chocolates of which 2 are filled with chilli and the rest are filled with caramel. If I pick r at random, show that the probability that I manage to avoid the chilli ones (n r)(n r 1) is. n(n 1) (vi) Show that n C r = n 1 C r + n 1 C r 1. This can be done either by manipulating factorials or by a clever choosing argument. We suggest that you try both ways. Note that this is the relationship that we see in Pascal s triangle: an entry in a given row is formed by adding together the two closest entries in the row above. (vii) The sequence a 0, a 1,..., a n has the property that a i /a i 1 > 1 for i k and a i /a i 1 < 1 for i > k. What is the largest term of the sequence? Statistics STEP Questions 15

15 The Probability question Bag P and bag Q each contain n counters, where n 2. The counters are identical in shape and size, but coloured either black or white. First, k counters (0 k n) are drawn at random from bag P and placed in bag Q. Then, k counters are drawn at random from bag Q and placed in bag P. If initially n 1 counters in bag P are white and one is black, and all n counters in bag Q are white, find the probability in terms of n and k that the black counter ends up in bag P. Find the value or values of k for which this probability is maximised. If initially n 1 counters in bag P are white and one is black, and n 1 counters in bag Q are white and one is black, find the probability in terms of n and k that the black counters end up in the same bag. Find the value or values of k for which this probability is maximised. 2015 S2 Q12 16 Preparation Suppose that we want to make a choice from 3 options, P, Q and R, but only have a fair coin to do this. One option is to toss the coin twice and then: if we get HH choose P, if we get HT choose Q, if we get TH choose R, and if we get TT then repeat, i.e. toss the coin twice again. (a) What is the probability that we choose P on the first two tosses of the coin? (b) What is the probability that we don t make a choice on the first two tosses, but choose P on the second two tosses? (c) Write down an infinite sum for the probability that we eventually choose P and hence find the probability that we choose P. Comment on your answer. This time we keep tossing the coin until we get one of the following sequences: if HH appears first, choose P, if HT appears first, choose Q, and if TH appears first, choose R. So if we have the sequence TTTTTH we would choose R as out of the three sequences, TH appears first. By considering what happens if the first toss is a Head or a Tail, find the probabilities that you choose P, Q and R. Statistics STEP Questions 16

17 The Probability question Four players A, B, C and D play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT. The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner. Show that, if only A and B play, then A has a probability of 1 4 of winning. If all four players play together, find the probabilities of each one winning. (iii) Only B and C play. What is the probability of C winning if the first two tosses are TT? Let the probabilities of C winning if the first two tosses are HT, TH and HH be p, q and r, respectively. Show that p = 1 2 + 1 2 q. Find the probability that C wins. Discussion It is hard to write a preparation for this question which does not give too much away. For the first part think carefully about who must win after the first two tosses in the 4 different cases. For the last part, you are asked to derive a given equation; then you need to find two more equations. Statistics STEP Questions 17