Probability Definitions

Showing 1-11 of 11 problems
2021 Paper 2 Q12
D: 1500.0 B: 1500.0

  1. A game for two players, \(A\) and \(B\), can be won by player \(A\), with probability \(p_A\), won by player \(B\), with probability \(p_B\), where \(0 < p_A + p_B < 1\), or drawn. A match consists of a series of games and is won by the first player to win a game. Show that the probability that \(A\) wins the match is \[ \frac{p_A}{p_A + p_B}. \]
  2. A second game for two players, \(A\) and \(B\), can be won by player \(A\), with probability~\(p\), or won by player \(B\), with probability \(q = 1 - p\). A match consists of a series of games and is won by the first player to have won two more games than the other. Show that the match is won after an even number of games, and that the probability that \(A\) wins the match is \[ \frac{p^2}{p^2 + q^2}. \]
  3. A third game, for only one player, consists of a series of rounds. The player starts the game with one token, wins the game if they have four tokens at the end of a round and loses the game if they have no tokens at the end of a round. There are two versions of the game. In the cautious version, in each round where the player has any tokens, the player wins one token with probability \(p\) and loses one token with probability \(q = 1 - p\). In the bold version, in each round where the player has any tokens, the player's tokens are doubled in number with probability \(p\) and all lost with probability \(q = 1 - p\). In each of the two versions of the game, find the probability that the player wins. Hence show that the player is more likely to win in the cautious version if \(1 > p > \tfrac{1}{2}\) and more likely to win in the bold version if \(0 < p < \tfrac{1}{2}\).

2016 Paper 1 Q12
D: 1516.0 B: 1484.7

  1. Alice tosses a fair coin twice and Bob tosses a fair coin three times. Calculate the probability that Bob gets more heads than Alice.
  2. Alice tosses a fair coin three times and Bob tosses a fair coin four times. Calculate the probability that Bob gets more heads than Alice.
  3. Let \(p_1\) be the probability that Bob gets the same number of heads as Alice, and let~\(p_2\) be the probability that Bob gets more heads than Alice, when Alice and Bob each toss a fair coin \(n\) times. Alice tosses a fair coin \(n\) times and Bob tosses a fair coin \(n+1\) times. Express the probability that Bob gets more heads than Alice in terms of \(p_1\) and \(p_2\), and hence obtain a generalisation of the results of parts (i) and (ii).

Show Solution
  1. There are several possibilities \begin{array}{c|c|c} \text{Alice} & \text{Bob} & P \\ \hline 0 & 1 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 2 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ 1 & 2 & 2 \cdot \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{6}{2^5} \\ 1 & 3 & 2\cdot \frac1{2^2} \cdot \frac{1}{2^3} = \frac{2}{2^5} \\ 2 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ \hline && \frac{1}{2^5}(3+3+1+6+2+1) = \frac{16}{2^5} = \frac12 \end{array}
  2. There are several possibilities \begin{array}{c|c|c} A & B & \text{count} \\ \hline 0 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 3 & 4 \\ 0 & 4 & 1 \\ 1 & 2 & 3\cdot6 \\ 1 & 3 & 3\cdot4 \\ 1 & 4 & 3 \\ 2 & 3 & 3\cdot4 \\ 2 & 4 & 3 \\ 3 & 4 & 1 \\ \hline && 64 \end{array} Therefore the total probability is \(\frac12\)
  3. \(\mathbb{P}(\text{Bob more than Alice}) = p_1 \cdot \underbrace{\frac12}_{\text{he wins by breaking the tie on his last flip}} + p_2\) If \(p_3\) is the probability that Alice gets more heads than Bob, then by symmetry \(p_3 = p_2\) and \(p_1 + p_2 + p_3 = 1\). Therefore \(p_1 + 2p_2 = 1\). ie \(\frac12 p_1 + p_2 = \frac12\) therefore the answer is always \(\frac12\) for all values of \(n\).
2013 Paper 1 Q12
D: 1500.0 B: 1468.0

Each day, I have to take \(k\) different types of medicine, one tablet of each. The tablets are identical in appearance. When I go on holiday for \(n\) days, I put \(n\) tablets of each type in a container and on each day of the holiday I select \(k\) tablets at random from the container.

  1. In the case \(k=3\), show that the probability that I will select one tablet of each type on the first day of a three-day holiday is \(\frac9{28}\). Write down the probability that I will be left with one tablet of each type on the last day (irrespective of the tablets I select on the first day).
  2. In the case \(k=3\), find the probability that I will select one tablet of each type on the first day of an \(n\)-day holiday.
  3. In the case \(k=2\), find the probability that I will select one tablet of each type on each day of an \(n\)-day holiday, and use Stirling's approximation \[ n!\approx \sqrt{2n\pi} \left(\frac n\e\right)^n \] to show that this probability is approximately \(2^{-n} \sqrt{n\pi\;}\).

Show Solution
  1. The probability the first is different to the second is \(\frac68\), the probability the third is different to both of the first two is \(\frac37\) therefore the probability is \(\frac{6}{8} \cdot \frac37 = \frac9{28}\) Whatever pills we are left with on the last day is essentially the same random choice as we make on the first day, therefore \(\frac9{28}\)
  2. The probability the first is different to the second is \(\frac{2n}{3n-1}\), the probability the third is different to both of the first two is \(\frac{n}{3n-2}\) therefore the probability is \(\frac{2n^2}{(3n-1)(3n-2)}\). [We can also view this as \(\frac{(3n) \cdot (2n) \cdot n}{(3n) \cdot (3n-1) \cdot (3n-2)}\)]
  3. Suppose describe the pills as \(B\) and \(R\) and also number them, then we must have a sequence of the form: \[ B_1R_1 \, B_2R_2 \, B_3R_3 \, \cdots \, B_{n}R_n \] However, we can also rearrange the order of the \(B\) and \(R\) pills in \(n!\) ways each, and also the order of the pairs in \(2^n\) ways. There are \((2n)!\) orders we could have taken the pills out therefore the probability is \begin{align*} && P &= \frac{2^n (n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}{n}} \\ &&&\approx \frac{2^n \cdot 2n \pi \left ( \frac{n}{e} \right)^{2n}}{\sqrt{2 \cdot 2n \cdot \pi} \left ( \frac{2n}{e} \right)^{2n}} \\ &&&= \frac{2^n \sqrt{n \pi} \cdot n^{2n} \cdot e^{-2n}}{2^{2n} \cdot n^{2n} \cdot e^{-2n}} \\ &&&= 2^{-n} \sqrt{n \pi} \end{align*} There is a nice way to think about this question using conditional probability. Suppose we are drawing out of an infinitely supply of \(R\) and \(B\) pills, then each day there is a \(\frac12\) chance of getting different pills. Therefore over \(n\) days there is a \(2^{-n}\) chance of getting different pills. Conditional on the balanced total we see that \begin{align*} && \mathbb{P}(\text{balanced every day} |\text{balanced total}) &= \frac{\mathbb{P}(\text{balanced every day})}{\mathbb{P}(\text{balanced total})} \end{align*} We have already seen the term that is balanced total is \(\frac{1}{2^{2n}}\binom{2n}{n}\), but we can also approximate the balanced total using a normal approximation. \(B(2n, \tfrac12) \approx N(n, \frac{n}{2})\) and so: \begin{align*} \mathbb{P}(X = n) &\approx \mathbb{P}\left (n-0.5 \leq \sqrt{\tfrac{n}{2}} Z + n \leq n+0.5 \right) \\ &= \mathbb{P}\left (- \frac1{\sqrt{2n}} \leq Z \leq \frac{1}{\sqrt{2n}} \right) \\ &= \int_{- \frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \d x \approx \frac{2}{\sqrt{2n}} \frac{1}{\sqrt{2\pi}} \\ &\approx \frac{1}{\sqrt{n\pi}} \end{align*}
2012 Paper 1 Q13
D: 1500.0 B: 1529.2

I choose at random an integer in the range 10000 to 99999, all choices being equally likely. Given that my choice does not contain the digits 0, 6, 7, 8 or 9, show that the expected number of different digits in my choice is 3.3616.

Show Solution
We are choosing any \(5\) digit number from \(\{1,2,3,4,5\}\). There are \(5^5\) such numbers. \begin{align*} && \mathbb{E}(\text{different digits}) &= \frac1{5^5} \left (1 \cdot 5 + 2 \cdot \binom{5}{2}(2^5-2)+3 \cdot \binom{5}{3}(3^5-3 \cdot 2^5+3)+4 \cdot \binom{5}{4}(4^5 - 4 \cdot 3^5+6 \cdot 2^5-4) + 5 \cdot 5! \right) \\ &&&= \frac{2101}{625} = 3.3616 \end{align*}
2011 Paper 1 Q12
D: 1500.0 B: 1470.2

I am selling raffle tickets for \(\pounds1\) per ticket. In the queue for tickets, there are \(m\) people each with a single \(\pounds1\) coin and \(n\) people each with a single \(\pounds2\) coin. Each person in the queue wants to buy a single raffle ticket and each arrangement of people in the queue is equally likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets if I cannot give the required change.

  1. In the case \(n=1\) and \(m\ge1\), find the probability that I am able to sell one ticket to each person in the queue.
  2. By considering the first three people in the queue, show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=2\) and \(m\ge2\) is \(\dfrac{m-1}{m+1}\,\).
  3. Show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=3\) and \(m\ge3\) is \(\dfrac{m-2}{m+1}\,\).

Show Solution
  1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
  2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
  3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows
2008 Paper 1 Q13
D: 1500.0 B: 1452.7

Three married couples sit down at a round table at which there are six chairs. All of the possible seating arrangements of the six people are equally likely.

  1. Show that the probability that each husband sits next to his wife is \(\frac{2}{15}\).
  2. Find the probability that exactly two husbands sit next to their wives.
  3. Find the probability that no husband sits next to his wife.

2007 Paper 1 Q12
D: 1500.0 B: 1484.0

  1. A bag contains \(N\) sweets (where \(N \ge 2\)), of which \(a\) are red. Two sweets are drawn from the bag without replacement. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.
  2. There are two bags, each containing \(N\) sweets (where \(N \ge 2\)). The first bag contains \(a\) red sweets, and the second bag contains \(b\) red sweets. There is also a biased coin, showing Heads with probability \(p\) and Tails with probability \(q\), where \(p+q = 1\). The coin is tossed. If it shows Heads then a sweet is chosen from the first bag and transferred to the second bag; if it shows Tails then a sweet is chosen from the second bag and transferred to the first bag. The coin is then tossed a second time: if it shows Heads then a sweet is chosen from the first bag, and if it shows Tails then a sweet is chosen from the second bag. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.

2004 Paper 1 Q14
D: 1500.0 B: 1488.1

Three pirates are sharing out the contents of a treasure chest containing \(n\) gold coins and \(2\) lead coins. The first pirate takes out coins one at a time until he takes out one of the lead coins. The second pirate then takes out coins one at a time until she draws the second lead coin. The third pirate takes out all the gold coins remaining in the chest. Find:

  1. the probability that the first pirate will have some gold coins;
  2. the probability that the second pirate will have some gold coins;
  3. the probability that all three pirates will have some gold coins.

Show Solution
  1. The first pirate will have some gold coins as long as the very first coin drawn is a gold coin, ie \(\frac{n}{n+2}\).
  2. The probability the second pirate will have some gold coins is the probability the two lead coins are separate. There are \((n+2)!\) ways to arrange the coins, and there are \((n+1)! \cdot 2\) ways to arrange the coins where they are together, therefore the probability is: \[ 1 - \frac{2(n+1)!}{(n+2)!} = 1 - \frac{2}{n+2} = \frac{n}{n+2} \]
  3. For all three pirates to have some gold coins we need the lead coins to be separate and not first or last. If we line up all \(n\) gold coins, there are \(n-1\) gaps between them we could place the \(2\) lead coins in, therefore \(\binom{n-1}{2}\) ways to place the lead coins with the restriction. Without the restriction there are \(\binom{n+2}{2}\) ways to choose where to put the coins, therefore \begin{align*} && P &= \frac{\binom{n-1}{2}}{\binom{n+2}{2}} \\ &&&= \frac{(n-1)(n-2)}{(n+2)(n+1)} \end{align*} [Notice this is clearly \(0\) if there are only \(1\) or \(2\) gold coins]
2001 Paper 1 Q14
D: 1500.0 B: 1516.8

On the basis of an interview, the \(N\) candidates for admission to a college are ranked in order according to their mathematical potential. The candidates are interviewed in random order (that is, each possible order is equally likely).

  1. Find the probability that the best amongst the first \(n\) candidates interviewed is the best overall.
  2. Find the probability that the best amongst the first \(n\) candidates interviewed is the best or second best overall.
Verify your answers for the case \(N=4\), \(n=2\) by listing the possibilities.

Show Solution
  1. The probability the best person falls in the first \(n\) is \(\frac{n}{N}\)
  2. The probability the best two people do not fall in the first \(n\) candidates is \begin{align*} && 1-P &= \frac{\binom{N-2}{n}}{\binom{N}{n}} \\ &&&= \frac{(N-2)(N-3)\cdots(N-2-n+1)}{n!} \frac{n!}{N(N-1)(N-2) \cdots (N-n+1)} \\ &&&= \frac{(N-n)(N-n-1)}{N(N-1)} \\ \Rightarrow && P &= 1- \frac{(N-n)(N-n-1)}{N(N-1)} \\ &&&= \frac{N(N-1) - N(N-1)+n(N-n-1)+Nn}{N(N-1)} \\ &&&= \frac{n(2N-n-1)}{N(N-1)} \end{align*}
If \(N = 4, n = 2\) the possibilities are, the best candidate can be first \(3!\) ways, or second \(3!\) ways, which is \(\frac{12}{24} = \frac{1}{2} = \frac{2}{4} = \frac{n}{N}\) so our formula works. In the case neither of the best two candidates are in the first half, the possibilities are \(3412, 3421, 4312, 4321\), ie \(\frac{4}{24} = \frac16\) chance, so the probability they are selected in the first \(n\) is \(\frac56\). our formula says it should be \(\frac{2 \cdot (2 \cdot 4 - 2 - 1)}{4 \cdot 3} = \frac{2 \cdot 5}{4 \cdot 3} = \frac56\) as desired.
1995 Paper 1 Q12
D: 1500.0 B: 1501.9

A school has \(n\) pupils, of whom \(r\) play hocket, where \(n\geqslant r\geqslant2.\) All \(n\) pupils are arranged in a row at random.

  1. What is the probability that there is a hockey player at each end of the row?
  2. What is the probability that all the hockey players are standing together?
  3. 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.

1991 Paper 1 Q14
D: 1516.0 B: 1457.1

A set of \(2N+1\) rods consists of one of each length \(1,2,\ldots,2N,2N+1\), where \(N\) is an integer greater than 1. Three different rods are selected from the set. Suppose their lengths are \(a,b\) and \(c\), where \(a > b > c\). Given that \(a\) is even and fixed, show, by considering the possible values of \(b\), that the number of selections in which a triangle can then be formed from the three rods is \[ 1+3+5+\cdots+(a-3), \] where we allow only non-degenerate triangles (i.e. triangles with non-zero area). Similarly obtain the number of selections in which a triangle may be formed when \(a\) takes some fixed odd value. Write down a formula for the number of ways of forming a non-degenerate triangle and verify it for \(N=3\). Hence show that, if three rods are drawn at random without replacement, then the probability that they can form a non-degenerate triangle is \[ \frac{(N-1)(4N+1)}{2(4N^{2}-1)}. \]

Show Solution
Suppose we have \(a = 2k\), it is necessary (by the triangle inequality) that \(b + c > a\). So the smallest \(b\) can be is \(k+1\), and then \(c\) must be \(k\) (1 choice). Then \(b\) could be \(k+2\) and \(c\) can be \(k+1\), \(k\), \(k-1\) (3 choices). Suppose \(b = k+i\) then \(c\) can be \(k+i-1, \ldots, k-i+1\) which is \(2i-1\) choices. This works until \(b = 2k-1\) and there are \(2(k-1)-1 = 2k-3 = a-3\) choices. Therefore there are \(1 + 3 + 5 + \cdots + (a-3)\) total choices. If \(a = 2k+1\) then, \(b = k+1\) is not possible \(b = k+2\) we have \(a = k+1, k\) (2 choices) \(b = k+3\) we have \(a = k+2, k+1, k, k-1\) (4 choices) \(b = k + i\) we have \(a = k+i-1, \cdots, k-i+2\) (\(2i-2\) choices) This works until \(b = k+k\) with \(2k-2 = a-3\) choices. So \(2 + 4 + \cdots + (a-3)\) If \(a\) is even, we have \(\left ( \frac{a-2}{2} \right)^2\) If \(a\) is odd we have \(\frac{(a-3)(a-1)}{4}\) Therefore the total number is: \begin{align*} C &= \sum_{k=2}^N \left ( \frac{(2k-2)^2}{4} + \frac{(2k+1-3)(2k+1-1)}{4} \right) \\ &= \sum_{k=2}^N \left ( (k-1)^2 + (k-1)k\right) \\ &= \sum_{k=2}^N (2k^2-3k+1) \\ &= \sum_{k=1}^N (2k^2-3k+1) \\ &= \frac{N(N+1)(2N+1)}{3} - \frac{3N(N+1)}{2} + N \\ &= \frac{N((N+1)(4N+2-9)+6)}{6} \\ &= \frac{N(4N+1)(N-1)}{6} \\ \end{align*} When \(N = 3\) we have \(1, 2, \cdots, 7\) sticks, and so \(a = 4\), \(1\) option \(a = 5\), \(2\) options \(a = 6\) \(4\) options \(a = 7\) \(6\) options for a total of \(13\). \(\frac{3 \cdot 13 \cdot 2}{6} = 13\) so this is promising, There are \(\binom{2N+1}{3}\) ways to choose three sticks (in order) and of those our formula tells us how many are valid, therefore \begin{align*} && P &= \frac{ \frac{N(4N+1)(N-1)}{6} }{\frac{(2N+1)2N(2N-1)}{6}} \\ &&&= \frac{(4N+1)(N-1)}{2(4N^2-1)} \end{align*}