There are 3 sets of balls numbered 1 through 10 placed in a bowl. If 3 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.

After the first ball is picked, there are 29 balls left, 2 of which have the same number as the first ball. If we pick two equal balls, there is now a single equal ball left out of 28 balls in the bowl. The probability of picking three equal balls is:

[tex]P=\frac{2}{29} *\frac{1}{28}\\P=\frac{1}{406}=0.0024630[/tex]

The probability is 1/406 or 0.0024630.

joaobezerra joaobezerra · Answer 2 · 2021-12-14T13:41:53+01:00

Using the hypergeometric distribution, it is found that there is a 0.002463 = 0.2463% probability that the balls have the same number.

The balls are chosen without replacement, hence, the hypergeometric distribution is used.

Hypergeometric distribution:

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

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

The parameters are:

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.

In this problem:

There are 30 balls, hence [tex]N = 30[/tex].

For each number, there are 3 balls, hence [tex]k = 3[/tex].

3 balls are selected, hence [tex]n = 3[/tex].

For each ball, the probability is P(X = 3). There are 10 balls, hence we have to find 10P(X = 3).

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

[tex]P(X = 3) = h(3,30,3,3) = \frac{C_{3,3}C_{27,0}}{C_{30,3}} = 0.0002463[/tex]

0.0002463 x 10 = 0.002463

0.002463 = 0.2463% probability that the balls have the same number.

For more on the the hypergeometric distribution, you can check https://brainly.com/question/24826394