Question

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.

a. What is the probability that an entire season elapses with a single no-hitter?

b. If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season?

c. What is the probability that there are more than $3$ no-hitters in a single season?

Short answer

a). The probability that an entire season elapses with a single no-hitter is $0.0498$.

b). The value of $P(\mathrm{T}>2\mid \mathrm{T}>1)$is $0.0498$.

c). The value of $P(x>3)$is $0.3528$.

Step-by-step solution

Part (a) Step 1: Given Information 

$X~Exp\left(3\right)$

Where.

$m=3$

$\mu =\frac{1}{3}$

$=0.3333$

Part (a) Step 2: Explanation

The required probability can be calculated as below:

$P(x>1)=1-P(x<1)$

$=1-\left(1-e^{-3-1}\right)$

$=e^{-3}$

$=0.0498$

Part (b) Step 1: Given Information 

$X~Exp\left(3\right)$

Where.

$m=3$

$\mu =\frac{1}{3}$

$=0.3333$

Part (b) Step 2: Explanation 

The required probability can be written as $P(\mathrm{T}>2\mid \mathrm{T}>1)$. Now, use the memory less property of exponential distribution. That is,

$P(x>t+r\mid x>t)=P(x>r)$

So,

$P(\mathrm{T}>2\mid \mathrm{T}>1)=P(T>1+1\mid \mathrm{T}>1)$

$=P(T>1)$

$=0.0498$

Therefore, the value of $P(\mathrm{T}>2\mid \mathrm{T}>1)$is $0.0498$.

Part (c) Step 1: Given Information 

$X~Exp\left(3\right)$

Where.

$m=3$

$\mu =\frac{1}{3}$

$=0.3333$

Part (c) Step 2: Explanation (Part c)

Let $X=$the number of no-hitters is a season. Assume that $X$is Poisson with mean $\lambda =3$. Then,

localid="1650470480426" $$\begin{gathered} P(X>3)=1-P(X\le 3) \\ =0.3528 \end{gathered}$$.