Problem of the week #49 - March 3rd, 2013

[Jameson]

Each individual in a group of n students is asked to pick an integer at random between 1 and 10 (inclusive). What is the smallest value of n that assures at least a 50% chance that at least two students select the same integer?

Show your work! Intuition is not enough for this problem and might be wrong.

Hint:

Spoiler

This problem is based off of the Birthday problem. You can use the same method to solve it.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Solution (from anemone):

Spoiler

$\displaystyle \text{ P(at least two students select the same integer)} \ge 0.5$

$\displaystyle1-\text{P(all of the students select different integer)} \ge 0.5$

$\displaystyle 1-0.5 \ge \text{P(all of the students select different integer)}$

$\displaystyle 0.5\ge \text{P(all of the students select different integer)}$

$\displaystyle 0.5 \ge \frac{10!}{(10-n)!(10^n)} $

Now, by making a two-column table for both the values for $n$ and $\displaystyle \frac{10!}{(10-n)!(10^n)} $ and start calculating from $n=1$, we find that the smallest value of n that assures at least a 50% chance that at least two students select the same integer occurs when $n=5$.