William P. answered • 07/19/19

Bill P., Math and Physics Tutor

Hello Greedy,

Define a random variable X to be the total number of hits, runs, and rbi's the player gets in any randomly selected game. Hits, runs, and rbi's could each be considered individual random variables, and they are certainly not independent, but we are still justified in using the data given to conclude that the expected value (or mean) of X is

μ = E[X] = 1.09 + .5 + .64 = 2.23

It is important to note that although the average value of X (per game) is a decimal number, the actual value of X in any one given game must be a whole number. That is, X is a** discrete random variable**.

The simplest method to approach the problem is to assume that X follows a **Poisson distribution**. A discrete random variable X has the Poisson distribution if its probability mass function is given by

(Eq.1) p(x) = (e^{-μ}μ^{x})/x!, for x = 0,1,2,3,4,......

In the above formula, the parameter μ is the mean value of X, and, as noted previously, in our problem, X is the number of hits + runs + rbi's in any given game. We want the probability that X meets or exceeds 1.5. That is, we want **P(X≥1.5)**. But since X takes on only whole number values, **this is the same P(X≥2)** (the probability that X is greater than or equal to 2.) That is,

(Eq.2) P(X≥1.5) = P(X≥2)

= P(2) + P(3) + P(4) + P(5) + ....

Since X≥2 corresponds, at least in theory, to infinitely many possible values of X, we would have to apply the Poisson formula infinitely many times. Instead of calculating the probability of X≥2 directly, calculate the probability of the **complement** of the event X≥2 and subtract this from 1. **The complement of "X≥2" is "X<2"**, which is equivalent to X = 0 or 1. Therefore,

(Eq.3) P(X≥2) = 1 - P(X<2)

= 1 - [p(0) + p(1)]

From the Poisson distribution given above (Eq.1), we have

p(0) = (e^{-2.23}·2.23^{0})/0!

**p(0) ≅ .1075**

and

p(1) = (e^{-2.23}·2.23^{1})/1!

**p(1) ≅ .2398**

Plug these results in to Eq.3 to obtain

P(X≥2) = 1 - [.1075 + .2398]

P(X≥2) = 1 - .3473

**P(X≥2) ≅ .6527** (rounded to the fourth decimal place.)

Therefore, the probability that X is at least 1.5 [the same as P(X≥2)] is **.6527**. Formally, any probability p is a number between 0 and 1, inclusive (0≤p≤1). However, it is often convenient to informally report a probability in percent terms, which in our case, would be **65.27%**.

Hope that helps! Please let me know if you need any additional explanation.

William

Greedy R.

Bill, I think you may be a genius ;) Thank you so much for your answer!!!! I was truly unable to solve this on my own, and your explanation is elegant and extremely detailed. I just recreated the equations that you shared, and I was finally able to solve for the answer. Not only that, but even more importantly I now understand the math behind it, and how to properly apply it. I really can't thank you enough, and truly appreciate the time, thoughtfulness, and expertise you provided! Cheers!07/20/19