O God? Is there's something wrong with the answers you've gotten?

Just to be pedantic, a "chance" is a number in the interval [0, 1]. "1 in 384,615" is a chance. "384,615" is not.The chance of reaching into the bucket and picking out a red ball are:

(20,000,000 / 52) = 384,615

Meaning that the probability of drawing out a red ball is 1 in 384,615 chances.

I'm not particularly a fan of pedantry, but let's do our utmost here to strive for clarity.

What does "confident" mean to you? This drawing will fail 36.79% of the time. This, like everything else you ask,Meaning that I would have to reach into the bucket 384,615 times in order to be confident of drawing out a red ball.has been explained in the thread already.

Wrong. Let's see if we can agree on WHY it's wrong.

First, humor us, and use probabilities that are between 0 and 1. IOW, write (52 * 20000) / 20000000 = 1/19.23. (Your introduction of "384,615 " is just a distraction.) BTW, 1/19.23 is anapproximationto theexactanswer 5.2000000%. Do you see that?

But 0.052 is NOT the probability you will draw exactly one red, nor is it the probability you will draw at least one red.It is the number of reds you will draw on average.

The probabilities you will draw exactly 0, 1, 2, 3, 4 or 5 reds are

p_{0}= .9493041051

p_{1}= .0494133528

p_{2}= .0012612419 ~ .05^2 / 2!

p_{3}= .0000210397 ~ .05^3 / 3!

p_{4}= .0000002580 ~ .05^4 / 4!

p_{5}= .0000000025 ~ .05^5 / 5!

(I've shown the probabilities along with an approximation, mentioned upthread, that may be useful for quick estimates.)

p_{1}is 0.049..., not 0.052. To get 0.052 you'll need the sum p_{1}+ 2*p_{2}+ 3*p_{3}+ 4*p_{4}+ ...

Does any of this help?

When we look at shades of red, please clarify the ambiguity identified above in #17 ( "(a) or (b)" ).If we can just settle that for the moment then we can look at shades of red later. OK?

cheers ... Greg