Binomial probability, similar to lottery problems.

[alexburns1991]

Homework Statement

An opaque bag contains 10 green counters, and 20 red ones. One counter is drawn at random and not replaced: green scores one, red scores zero. Five counters are drawn.

Find the probability of scoring 0, 1, 2, 3, 4, 5 points.

Homework Equations

The Attempt at a Solution

I found it pretty straighforward to work out with replacement, as it is just the simple binomial probability. But when the counters aren't replaced, surely the order counts, so i tried replacing nCr with nPr though this gave me the completely wrong answer. i know this resembles the lottery problem, but i don't understand what to do when there are only 2 distinguishable groups.

 

[vela]

If you score n points, that means you drew n green and 5-n red counters. What you want to do is figure out the number of ways you can do that and divide by the total number of ways you can draw five items from thirty.

 

[Architeuthis Dux]

The binomial probability formula can still be used in this situation, even without replacement. The only difference is that the probability of success (drawing a green counter) will change with each draw, as there are fewer counters in the bag after each draw.

To find the probability of scoring 0 points, you would use the formula P(0) = (20/30) * (19/29) * (18/28) * (17/27) * (16/26) = 0.2334. This is because with each draw, the probability of drawing a red counter increases slightly.

Similarly, to find the probability of scoring 1 point, you would use the formula P(1) = (10/30) * (20/29) * (19/28) * (18/27) * (17/26) = 0.4017. And for 2 points, P(2) = (10/30) * (9/29) * (20/28) * (19/27) * (18/26) = 0.2527.

The rest of the probabilities can be calculated in the same way. The key is to adjust the probability of success for each draw, based on the number of counters left in the bag.