Question

There are 88 black balls and 88 red balls in an urn. If 33 balls are drawn without replacement, what is the probability that exactly 11 black ball is drawn? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Total of 88 + 88 = 176 balls, so $N = 176$

33 balls are drawn, so $n = 33$

We want 11 black balls(sucesses), so $n = 11$

There are 88 black balls, so $k = 88$

Then

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 11) = h(11,176,88,33) = \frac{C_{88,11}*C_{88,22}}{C_{176,33}} = 0.0164$

0.0164 probability that exactly 11 black balls are drawn