I need some help on stats. So there are 100 balls, r of which are red. What is the probability that the 50th and last balls drawn will be red, without replacement. Thanks.
I need some help on stats. So there are 100 balls, r of which are red. What is the probability that the 50th and last balls drawn will be red, without replacement. Thanks.
P = (r/100)*((r-1)/100)
So if there are 50 red out of 100, the chance that the 50th draw is red is 50%. The chance that any subsequent draw is red is 49%. The chance that the 50th and any other draw are both red is 24.5%.
Before any balls are drawn, the probability of a red ball in any of the 100 single draws is r/100. So if that's the question, the probability is (r/100)*(r/100). If you work out a small number of draws, say 2 red out of 4 draws, you will see that the probability remains .5 across all of the possible scenarios for each draw.
I think it would be (r/100)*((r-1)/99). Once you've removed one red the probability of drawing the last one as red has to reflect this on both the numerator and denominator.
probably not wrote:
Before any balls are drawn, the probability of a red ball in any of the 100 single draws is r/100. So if that's the question, the probability is (r/100)*(r/100). If you work out a small number of draws, say 2 red out of 4 draws, you will see that the probability remains .5 across all of the possible scenarios for each draw.
This is a dependent probability problem not an independent one, much like calculating poker probabilities.
The formula is
P = (r/100)*((r-1)/99) = 24.74%
Since you brought up the 2 out of 4 draw analogy lets run the calculation, then do a manual experiment.
P = (2/4)*(1/3) = 2/12
Manual proof. 4 balls: 2 Red (Rr) and two Blue (Bb)
R r B b
There are 12 possible combinations.
2 red/red
8 red/blue
2 blue/blue
Donrad wrote:
I think it would be (r/100)*((r-1)/99). Once you've removed one red the probability of drawing the last one as red has to reflect this on both the numerator and denominator.
Beat me to it.
I read the problem as being: what is the probability of picking a red ball on the 50th pick and picking a red ball on the 100th pick.
how are you able to come up with a probability of 24.74% if you do not know what r is to start with, especially since it is contingent on all the intermediate draws?
You guys are good at math. Man, I'm so dumb. :-(
They are assuming that R=50
Hey malmo what's your job. Just wondering as you seem to be an expert on most subjects, from dew point to complex maths.Not ass-licking, just wondering.
From what I have, it is the 50th row in Pascal's triangle which contains 50 terms, all of which are weighted differently. Not knowing the original red ball count leaves this essentially open ended in my opinion as there is no where to start.
The first correct answer.
Most people are not reading the original question correctly.
Because it is stated that there is no replacement of the balls, and that we are inerested in exactly the outcome of the 50th pick and the last pick (100th), and since we do not know how many red balls there are to begin with, there is no numerical answer.
how? wrote:
how are you able to come up with a probability of 24.74% if you do not know what r is to start with, especially since it is contingent on all the intermediate draws?
You are correct. My bad. I worked out an example with 50 red balls and 100 draws.
Here is the formula the OP was looking for.
P = (r/d)*[(r-1)/(d-1)]
Waloddi wrote:
I read the problem as being: what is the probability of picking a red ball on the 50th pick and picking a red ball on the 100th pick.
Yes it is. But whether or not the draws fall on those ordered draws (50th, 100th) exactly isn't any different than any two other ordered draws (1st, 2nd).
There is a difference when the balls are not being replaced.
Think of the decision tree and how it would alter depending on the previous outcomes.
The first draw you have the following probability:
(r/100) of choosing red and ((100-r)/100) of choosing not red.
The probability of drawing red on the second draw depends on whether the first ball drawn was red or not. The second draw of the decision tree looks like this if a red ball was chosen first:
((r-1)/99) ((100-r)/99)
If a non-red ball was chosen first you have:
(r/99) ((99-r)/99)
So, two red balls drawn in a row without replacement:
(r/100)*((r-1)/99)
If a non-red ball was chosen first followed by a red ball, without replacement:
((100-r)/100)*(r/99)
The path you take down the decision tree depends on the color of the previous ball chosen. The answer will be a product of 50 fractions that represent the chance of drawing that red ball given the 49 previous selections.
precisely wrote:
There is a difference when the balls are not being replaced.
Think of the decision tree and how it would alter depending on the previous outcomes.
I'm not sure what you are trying to say here? Previous outcomes are irrelevant. You are not making decisions during a string of outcomes, you are determining the probability beforehand. Whether or not the balls in question are the 50th and 100th, or the 1st and 2nd, or the 99th and 100th balls is irrelevant.
The formula is P = (r/d)*[(r-1)/d-1]
Lets suppose the question was: "So there are 3 balls, 2 of which are red. What is the probability that the 1st and last balls drawn will be red, without replacement."
Lets break it down manually for you again and compare it to the formula. This time there are three balls: 2 red and 1 Blue. I'm going to call the red ones X and x, the blue one O (I chose X and O becasue it's easier to see)
There are 6 possible orders of balls:
X x O
X O x
x X O
x O X
O x X
O X x
Manually you will see that there are 6 possible results: XO, Xx, xO, xX, OX, Ox
2 of 6 (33.3%) possible combinations will give you two reds.
How does that reconcile with the formula P = (r/d)*[(r-1)/d-1]?
P = (2/3)*(1/2) = 33.3% This is not a coincidence!
It doesn't matter which how you contruct the question the answer is the same.
"So there are 3 balls, 2 of which are red. What is the probability that the 1st and 2nd balls drawn will be red, without replacement." The answer is still 33.3%
"So there are 3 balls, 2 of which are red. What is the probability that the 2nd and 3rd balls drawn will be red, without replacement." The answer is still 33.3%
Copy or memorize this quote:
"in no other branch of mathematics is it so easy for experts to blunder as in probability theory" - Martin Gardner
Malmo - I think what precisely was (not intentionally) doing was drawing out all the possible combinations. Think about it this way using your example - there are 6 possible paths. In the example with 100 draws there are 100! possible paths (please correct me if i'm wrong with that one, ha! this is, of course, considering each ball as uniquely numbered 1-100 - regardless of color). Now those "possible" paths occur and each path has a probability of 1/100! of occurring (again, keep in mind that this considers each ball numbered 1-100, not colored). So it is path 1 OR path 2 OR path three.... which, if they have a red ball at 50 and 100, should add up to the correct answer: (r/d)*(r-1)/(d-1).
As you said - the confusion sets in when you think too hard. It is always too simple to simplify a problem. But it always ends up being the right way to do it. Did you know that if you have 2 red balls that there is a 0.0204% chance of picking them 50th and 100th in no particular order while not replacing any balls?
Also, to elaborate on the "replacement" issue - if you DID replace the balls after placing them it is a simple probability of (r/d)*(r/d) of picking a red ball at the 50th and the 100th draw. I love this stuff, don't know why. But wish I remembered more from my college courses.
sorry, finger slipped - 0.0202% chance of picking the 50th and 100th ball red when there are 2 red balls in the bunch.