A box contains 65 balls numbered from 1 to 65. If 11 balls are drawn with replacement, what is the probability that at least two of them have the same number?

Respuesta :

Using the binomial distribution, there is a 0.265 = 26.5% probability that at least two of them have the same number.

What is the binomial distribution formula?

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, the values of the parameters are:

p = 1/65 = 0.0154, n = 65.

The probability that at least two of them have the same number is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which:

P(X < 2) = P(X = 0) + P(X = 1)


Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{65,0}.(0.0154)^{0}.(0.9846)^{65} = 0.3647[/tex]

[tex]P(X = 1) = C_{65,1}.(0.0154)^{1}.(0.9846)^{64} = 0.3703[/tex]

So:

P(X < 2) = P(X = 0) + P(X = 1) = 0.3647 + 0.3703 = 0.735.

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.735 = 0.265[/tex]

0.265 = 26.5% probability that at least two of them have the same number.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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