In order to get a mean of 4/6, you would need to pick four 1's and two 0's. The probability of picking a 1 (with replacement) is 6/10 each time and probability of picking a 0 (with replacement) is 4/10. So,
(6/10)*(6/10)*(6/10)*(6/10)*(4/10)*(4/10)
However, you can pick the four 1's and two 0's in any order. So, you perform
binomial expansion
N!/[K!(N-K)!]= number of possible ways that K individuals can be selected from N
10!/[4!(10-4)!]
Finally, take the product.
10!/[4!(10-4)!]*(6/10)*(6/10)*(6/10)*(6/10)*(4/10)*(4/10)=?
Given the population:
0,0,0,0,1,1,1,1,1,1
If a sample size of 6 was taken at random with replacement, what is the probability that the sample mean is 4/6?
1 answer