Expected Value and Probability distribution - HELP I'm stuck!?

Question

I have a total of 1000 marbles, with 300 red, 600 blue, 100 green.

P(red marble) = 30%
P(blue marble) = 60%
P(green marble) = 10%

If I take a sample of 10 marbles..

What is the probability distribution of red marbles marbles? In other words, what is the expected value of getting 0,1,2,3,4,5,6,7,8,9,10 red marbles in the sample?

I don't need a list of the E(x) of 0-10, but please show work on how to do one or two examples.

Answer 1

This is a binomial distribution with p = 0.30
The general formula is:
nCx*p^x(1-p)^(n-x), where nCx = n!/[x!(n-x)!]
For example, the probability of getting 0 red marbles in a sample of 10:
P(X = 0) = 1*0.30^0(1-0.30)^10 = 0.0283
The probabaility of getting exactly 4 red marbles in a sample of 10:
P(x = 4) = 10!/[4!(10-4)!]0.30^4(1-0.30)^(10-4) = 0.2001

Answer 2

A freaky die. you're able to write the threat distribution this kind P(-a million) = a million/10 P(-2) = a million/10 P(-3) = a million/10 P(a million) = 7/60 P(2) = 14/60 P(3) = 21/60 That provides to a million, perfect? Your instructor in all probability needs For x < 0, P(x) = a million/10 For x > 0, P(x) = 7x/60 (actual, your instructor in all probability made a mistake, and became thinking that the entire threat for the useful numbers is 6/10. you're taking your opportunities featuring the incredible answer. some instructors do in comparison to pupils getting the incredible answer while they themselves are incorrect.)

Answer 3

prob(4 red, 6 not red) = [10C4][0.3^4][0.7^6] = 210 • 0.0081 • 0.17649 = 0.009529569

prob(a red, 10-a not red) = [10Ca][0.3^a][0.7^(10-a)]