# An urn problem: few red balls, many draws (with replacement)

So, this is a freshman probability problem and I am embarrassed to p[ost it, but I have been up for 35 hours and my brain is broken.

I have an urn with 60,000 white balls and 6 red balls. From this urn I make 60,000 draws with replacement. What is the probability that I draw no red balls?

Doing this numerically, first with a spreadsheet and then with R, I keep getting 0.00248, roughly. ((1- 6/60,000)^60,000) I expected it to be small, but that seems too small, intuitively, Also, the number is suspiciously close to e/1000, which kind-of fits, I think, raising a really low probability to a really high power. But the answer is on the order of 100 times smaller than I expect, based on just gut feeling.

Just for human interest, this number should be roughly the probability that the Current Population Survey sampled no one with incomes above $2 million in 2011, as indeed it didn’t. And assuming (falsely) that rich people are as likely to answer as anyone else. • The answer is$(60000/60006)^{60000}= .0024795\$...you were correct. Apr 2 '18 at 2:55
• Please add the self-study tag. Apr 2 '18 at 8:28
• @Xi'an ¿why does this require the self-study tag? The context of the problem has been stated in the question. Apr 2 '18 at 13:13
• Thanks @GregH! What I find confusing about this is that, while I think I ought to see six red balls a year, on the average, I don't see how, or why, the variance of that number can be so small. I'd expect to see no red balls or 12 red balls every decade or three, not once in 400 years. I suppose it is unreasonable to ask someone else to explain my faulty intuitions, but I wonder if you have any idea where I might be going off track? Apr 2 '18 at 17:15