Probability Distribution of 4 Balls: Mean & Variance

The probability of 1 red ball is (1/2)(3/7)(2/3)(1/5)+ (1/2)(4/7)(1/3)(1/5)= 3/35. The probability of 2 red balls is (1/2)(3/7)(2/3)(4/5)+ (1/2)(4/7)(1/3)(3/5)= 24/70= 12/35. The probability of 3 red balls is (1/2)(3/7)(1/3)(4/5)= 12/105. The probability of 4 red balls is (1/2)(3/7)(
  • #1
blumfeld0
148
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?
 
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  • #2
Do you know anything about binomial probability distribution ?
 
  • #3
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.
 

FAQ: Probability Distribution of 4 Balls: Mean & Variance

What is the probability distribution of 4 balls?

The probability distribution of 4 balls refers to the likelihood of each possible outcome occurring when four balls are randomly selected from a set. This can be represented by a table or graph that shows the probability of each outcome.

What is the mean of a probability distribution of 4 balls?

The mean, also known as the average, of a probability distribution of 4 balls is the sum of all possible outcomes multiplied by their respective probabilities. In simpler terms, it represents the expected value of the distribution.

How is the mean calculated for a probability distribution of 4 balls?

To calculate the mean of a probability distribution of 4 balls, you would multiply each possible outcome by its probability and then sum up all of these values. This gives you the expected value of the distribution.

What is the variance of a probability distribution of 4 balls?

The variance of a probability distribution of 4 balls measures the spread or variability of the distribution. It is calculated by finding the difference between each outcome and the mean, squaring these differences, and then multiplying them by their respective probabilities and summing them up.

How is the variance calculated for a probability distribution of 4 balls?

To calculate the variance of a probability distribution of 4 balls, you would first find the mean of the distribution. Then, for each possible outcome, you would subtract the mean, square the result, and multiply it by the probability. These values are then summed up to give you the variance.

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