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2015 Paper 1 Q12
D: 1500.0 B: 1461.6
probability Poisson distribution binomial distribution Poisson thinning conditional probability algebraic proof

The number \(X\) of casualties arriving at a hospital each day follows a Poisson distribution with mean 8; that is, \[ \P(X=n) = \frac{ \e^{-8}8^n}{n!}\,, \ \ \ \ n=0, \ 1, \ 2, \ \ldots \ . \] Casualties require surgery with probability \(\frac14\). The number of casualties arriving on any given day is independent of the number arriving on any other day and the casualties require surgery independently of one another.

  1. What is the probability that, on a day when exactly \(n\) casualties arrive, exactly \(r\) of them require surgery?
  2. Prove (algebraically) that the number requiring surgery each day also follows a Poisson distribution, and state its mean.
  3. Given that in a particular randomly chosen week a total of 12 casualties require surgery on Monday and Tuesday, what is the probability that 8 casualties require surgery on Monday? You should give your answer as a fraction in its lowest terms.

Solution:

  1. \(\mathbb{P}(r \text{ need surgery}|n \text{ casualties}) = \binom{n}{r} \left ( \frac14\right)^r \left ( \frac34\right)^{n-r}\)
  2. \(\,\) \begin{align*} && \mathbb{P}(r \text{ need surgery}) &= \sum_{n=r}^{\infty} \mathbb{P}(r \text{ need surgery} |n \text{ casualties}) \mathbb{P}(n \text{ casualties}) \\ &&&= \sum_{n=r}^{\infty} \binom{n}{r}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \sum_{n=r}^{\infty} \frac{n!}{(n-r)!r!}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{8^{n-r}}{(n-r)} \left ( \frac34\right)^{n-r} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{6^{n-r}}{(n-r)} \\ &&&= \frac{e^{-8}2^r}{r!} e^6 \\ &&&= \frac{e^{-2}2^r}{r!} \end{align*} Therefore the number requiring surgery is \(Po(2)\) with mean \(2\).
  3. \(\,\) \begin{align*} && \mathbb{P}(X_1 = 8| X_1 + X_2 =12) &= \frac{\mathbb{P}(X_1 = 8,X_2 =4)} {\mathbb{P}(X_1+X_2 = 12)}\\ &&&= \frac{\frac{e^{-2}2^8}{8!} \cdot \frac{e^{-2}2^4}{4!}}{\frac{e^{-4}4^{12}}{12!}} \\ &&&= \frac{12!}{8!4!} \frac{1}{2^{12}} \\ &&&= \binom{12}4 \left ( \frac12 \right)^4\left ( \frac12 \right)^8 \\ &&&= \frac{495}{4096} \end{align*}

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2009 Paper 1 Q12
D: 1500.0 B: 1501.5
probability tree diagram inequality AM-GM inequality combinatorics algebraic proof symmetric functions

Prove that, for any real numbers \(x\) and \(y\), \(x^2+y^2\ge2xy\,\).

  1. Carol has two bags of sweets. The first bag contains \(a\) red sweets and \(b\) blue sweets, whereas the second bag contains \(b\) red sweets and \(a\) blue sweets, where \(a\) and \(b\) are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot exceed the probability that they are of different colours.
  2. Simon has three bags of sweets. The first bag contains \(a\) red sweets, \(b\) white sweets and \(c\) yellow sweets, where \(a\), \(b\) and \(c\) are positive integers. The second bag contains \(b\) red sweets, \(c\) white sweets and \(a\) yellow sweets. The third bag contains \(c\) red sweets, \(a\) white sweets and \(b\) yellow sweets. Simon shakes the bags and picks one sweet from each bag without looking. Show that the probability that exactly two of the sweets are of the same colour is \[ \frac {3(a^2b+b^2c+c^2a+ab^2 + bc^2 +ca^2)}{(a+b+c)^3}\,, \] and find the probability that the sweets are all of the same colour. Deduce that the probability that exactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour.

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