Answer:
0.0164 probability that exactly 11 black balls are drawn
Step-by-step explanation:
The balls are drawn without replacement, so we use the hypergeometric distribution to solve this question.
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
In this question:
Total of 88 + 88 = 176 balls, so ![N = 176](https://tex.z-dn.net/?f=N%20%3D%20176)
33 balls are drawn, so ![n = 33](https://tex.z-dn.net/?f=n%20%3D%2033)
We want 11 black balls(sucesses), so ![n = 11](https://tex.z-dn.net/?f=n%20%3D%2011)
There are 88 black balls, so ![k = 88](https://tex.z-dn.net/?f=k%20%3D%2088)
Then
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![P(X = 11) = h(11,176,88,33) = \frac{C_{88,11}*C_{88,22}}{C_{176,33}} = 0.0164](https://tex.z-dn.net/?f=P%28X%20%3D%2011%29%20%3D%20h%2811%2C176%2C88%2C33%29%20%3D%20%5Cfrac%7BC_%7B88%2C11%7D%2AC_%7B88%2C22%7D%7D%7BC_%7B176%2C33%7D%7D%20%3D%200.0164)
0.0164 probability that exactly 11 black balls are drawn