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In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. What is the probability that an entire season elapses with a single no-hitter? If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? What is the probability that there are more than 3 no-hitters in a single season?

Respuesta :

Using the Poisson distribution, it is found that:

  • There is a 0.0498 = 4.98% probability that an entire season elapses with a single no-hitter.
  • If an entire season elapses without any no-hitters, there is a 0.0498 = 4.98% probability that there are no no-hitters in the following season.
  • There is a 0.3528 = 35.28% probability that there are more than 3 no-hitters in a single season.

What is the Poisson distribution?

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • [tex]\mu[/tex] is the mean in the given interval.

The average rate is of 3 no-hitters per season, hence:

[tex]\mu = 3[/tex].

The probability that an entire season elapses with a single no-hitter is P(X = 0), hence:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3}3^{0}}{(0)!} = 0.0498[/tex]

There is a 0.0498 = 4.98% probability that an entire season elapses with a single no-hitter.

Seasons are independent, hence:

If an entire season elapses without any no-hitters, there is a 0.0498 = 4.98% probability that there are no no-hitters in the following season.

The probability that there are more than 3 no-hitters in a single season is P(X > 3) given as follows:

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

Then:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3}3^{0}}{(0)!} = 0.0498[/tex]

[tex]P(X = 1) = \frac{e^{-3}3^{1}}{(1)!} = 0.1494[/tex]

[tex]P(X = 2) = \frac{e^{-3}3^{2}}{(2)!} = 0.2240[/tex]

[tex]P(X = 3) = \frac{e^{-3}3^{3}}{(3)!} = 0.2240[/tex]

Then:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472[/tex]

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.6472 = 0.3528[/tex]

There is a 0.3528 = 35.28% probability that there are more than 3 no-hitters in a single season.

More can be learned about the Poisson distribution at https://brainly.com/question/13971530

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