There are 4 sets of balls numbered 1 through 10 placed in a bowl. If 4 balls are randomly chosen without replacement, find the probability that the balls have the same number. Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Respuesta :

Solution:

Since the balls are chosen without replacement, we can apply the hypergeometric distribution. This distribution is given by the following formula:

[tex]P(X=x)=h(x,N,n,k)=\frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

where

[tex]C_{n,x}=\frac{n!}{x!(n-x)!}[/tex]

and the parameters are the following:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Now, in this problem we have that:

There are 4 sets of balls numbered 1 through 10 placed in a bowl. This means 4 x 10 = 40 balls, hence N= 40.

On the other hand, for each number, there are 4 balls, hence k = 4.

Finally, we have that 4 balls are selected, thus n= 4.

For each ball, the probability is P(X=4). There are 10 balls, hence we have to find 10 * P(X = 4).

Now, according to the given distribution, we obtain that:

[tex]P(X=4)=h(4,40,4,4)=\frac{C_{4.4}C_{40-4,4-4}}{C_{40,4}}[/tex]

this is equivalent to:

[tex]P(X=4)=h(4,40,4,4)=\frac{C_{4.4}C_{36,0}}{C_{40,4}}[/tex]

Now, applying the combinatorial formula to the above equation, we obtain:

[tex]P(X=4)=\frac{C_{4.4}C_{36,0}}{C_{40,4}}=\frac{1\cdot\text{ }1}{91390}=\frac{1}{91390}=0.00001094[/tex]

Thus,

[tex]10\cdot\text{ 0.00001094=}0.0001094[/tex]

in percent notation, this is equivalent to:

[tex]0.0001094\cdot100\%=0.01094\%[/tex]

So that, we can conclude that the correct answer is:

[tex]0.01094\%[/tex]