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STEP II 2007 question 13 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 13 solution

|\frac{r}{n}| is much smaller than 1 so we can assume \left( -\frac{r}{n} \right)^2 and higher powers are negligible.

\displaystyle e^{-r/n} = \sum_{k=0}^{\infty} \frac{\left( -\frac{r}{n}\right)^k}{k!} \approx 1 - \frac{r}{n} = \frac{n-r}{n}

The required probability is the same as 1-P(nobody shares a birthday). Which is \displaystyle 1 - \prod_{i=0}^{k-1} \frac{365-i}{365} \approx 1 - \prod_{i=0}^{k-1} e^{-i/365} = 1-e^{- k(k-1)/730} where we have used \displaystyle \sum_{i=0}^{k-1} i = \frac{1}{2}k(k-1).

Want the smallest natural number k such that

1-e^{-k(k-1)/730} \ge \frac{1}{2}

Notice that 22*23=506 = 2*253. So putting k=23 we have

1-e^{-506/730} = 1-e^{-253/365} = 1-\frac{1}{2} = \frac{1}{2}

using the approximation for log(2) given.

Also, since 1-e^{-k(k-1)/730} is strictly increasing for k in the natural numbers, k=23 is the smallest satisfying the inequality.

Suppose there are k guests. P(somebody having the host's birthday) = 1 - P(nobody has it) which is

1 - \left( \frac{364}{365} \right)^k = 1 - \left( \frac{365-1}{365} \right)^k \approx 1-e^{-k/365}

For an approximation of k we solve 1-e^{-k/365} \ge \frac{1}{2}

For k=253 we have equality. By a similar argument to the previous part, this is the smallest natural number satisfying this inequality.

Solution by SsEe.

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