I am struggling with a question as follows,
There will be a draw between 10000 attendants lets say. In total, 1000 people will be choosen. Me and 4 of my friends have joined to the draw. So we have 5 people in. I wonder, what should be my approach to calculate what is the probability that at least one of us will be choosen?
I mean, it looks like each have 1000/10000 = 0.1 probability to be choosen. But I cannot simply add 0.1 5 times because then if we were 10 people, result would be 100% probability that at least one of us will be choosen by this method, however, it is not true of course.
Also, another thing bothers me is the fact that the second one to be choosen will not be 0.1, instead 999/9999, and 3rd one 998/9998, and so on. So I cannot say that each will have 0.1 probability. If the numbers were smaller, the difference in probabilities would be very noticable.
As seen, I need a bit of enlightment about how to approach such a problem.
By the way, my best solution to this is
1 – (1 – 0.1)^5
Which is 1 – the joint probability of all of us not being choosen, which I think should give the probability that at least one of us will be chosen. However, the thing about 0.1 not being valid for 2nd, 3rd, 4th, and 5th attendants bothers me with this solution.
Thanks!
Best Answer
You are correct! This is actually because the two events, "A being chosen", and "B being chosen", are not disjoint. The only rule that allows you to add probabilities together is that if $A$ and $B$ are disjoint, then $P(A\cup B)=P(A)+P(B)$, but that rule is useless to you.
To get the correct answer, it's best to just count the number of outcomes.
The number of all outcomes is easy to calculate, it is simply ${10000\choose 1000}$. That's a pretty big number, but it is possible to calculate it.
The number of good outcomes is tough, but as you noted, it is a good idea to look at the opposite, the number of bad outcomes, in a sense. Then we can either
Either way, if none of the 5 people was chosen, then that means all $1000$ winners were chosen out of the set of $9995$ other possible winners, and there are ${9995\choose 1000}$ ways to do that, which means the probability of none of the $5$ being selected is
$$\frac{{9995\choose 1000}}{{10000\choose 1000}}$$
which simplifies to
$$\frac{9000\cdot 8999\cdots 8996}{10000\cdot 9999\cdots 9996}\approx 0.5904$$
Note that your solution, $1-(1-0.1)^5$, would be correct if the winners were chosen with replacement.