Probability Distribution of 4 Balls: Mean & Variance

AI Thread Summary
The discussion focuses on calculating the probability distribution, mean, and variance of drawing 4 balls from an urn containing 4 red and 4 white balls. It highlights the importance of determining whether the sampling is with or without replacement, as this significantly affects the probabilities. For sampling with replacement, the probability of drawing a red ball is 1/2, while for sampling without replacement, the probabilities change with each draw. The conversation also touches on the binomial probability distribution, emphasizing the need to account for different combinations when calculating probabilities for varying outcomes of red balls. Understanding these concepts is crucial for accurately setting up the probability distribution and computing the mean and variance.
blumfeld0
Messages
146
Reaction score
0
an urn contains 4 red balls and 4 white balls
an experiment consists of selecting at random a sample of 4 balls and
recording the number of red balls in the sample
setup the probability distribution and compute its mean and variance

i know what a probablity distribution is. can someone please how to calculate A probability and i can calculate the rest
also i know mean = total * probability of success = what numbers exactly?

variance = SQRT(total*sucsess * failure) = what numbers exactly?
 
Physics news on Phys.org
Do you know anything about binomial probability distribution ?
 
The first thing we need to know is whether this is "sampling with replacement" or "sampling without replacement". That is, whether a ball, after it is drawn from the urn and its color recorded is or is not returned to the urn.

Obviously, there are 5 possible outcomes: 0, 1, 2, 3, or 4 red balls in the sample.

Assuming "sampling with replacement", the probability of drawing a red ball is 4/8= 1/2 and the probability of drawing a white ball is 1/2 on each draw. In order to get 0 red balls, you have to draw a white ball each time: the probability of that is (1/2)(1/2)(1/2)(1/2)= 1/16. In order to get exactly 1 red ball, you will also need to consider the different orders in which it can be done: "red, white, white, white", "white, red, white, white", etc. That's where the binomial coefficient and the binomial probability distribution arnbg mentioned comes in.

If, on the other hand, this is "sampling without replacement", it's a lot harder! The probability of getting a white ball on the first draw is still 1/2 but IF that happens, there are now 4 red and 3 white balls in the urn. The probability of getting a second white ball is now 3/7. Then there are 4 red and 2 white balls in the urn. The probability of a third white ball is 2/6= 1/3. Finally, there are 4 red and only 1 white ball in the urn. The probability of drawing a fourth white ball is 1/5. The probability of 0 red balls (4 white balls) is (1/2)(3/7)(1/3)(1/5)= 1/70.
 
Back
Top