Question

expected value and variance for the described distribution? 5. Suppose that a box contains five red balls and ten blue balls. If seven balls are selected at random without replacement, what is the probability that at least 4 red balls will be obtained? Let X denote the proportion of red balls in the sample what are the mean and variance of X?

Answer 1

Solution:
Given in the question
In a box 5 red balls and 10 blue balls are given
Total population size (N) = 10+5 = 15
Sample size (n)= 7
No. of red in balls in the population (A) = 5
No. of red balls in the sample(x) = 4
And we are selecting 7 balls without replacement so we will use hypergeometric distribution to calculate the probability
we need to calculate the probability that at least 4 red balls will be obtained
P(X = x |n, N,A) = ACx * ((N-A)C(n-x))/NCn
P(X >=4 |7,15,5) = P(X = 4 |7,15,5) + P(X = 5 |7,15,5) = 5C4*((15-5)C(7-4))/15C7 + 5C5*((15-5)C(7-5))/15C7 = 0.0932 + 0.007 = 0.1002
So there is 10.02% probability that  at least 4 red balls will be obtained.
Mean of the distribution can be calculated as
Mean = (n*A)/N = (7*5)/15 = 2.33
the variance of distribution can be calculated as
Variance = (nA*(N-A)/N^2) * ((N-n)/(N-1)) = (7*5*(15-5)/(15*15)) * ((15-7)/(15-1)) = (350/225) * (8/14) = 2800/3150 = 0.89