Question

If the Mets begin this year as they have each of the past few years, they could end April with a record of 18 wins and 6 losses. If that trend were to continue, how many games could they be expected to

asked Nov 3, 2021 216k views

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Sisiutl · Answer 1 · 2021-11-07T06:57:58+0000

Answer:

p =(X)/(n)

Where:

X=18 number of wins in april

n = 18+6 = 24 total of games for April

And for this case we have:

p =(18)/(24)=0.75

$X \sim Binom(n=162, p=0.75)$

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^(n-x)

Where (nCx) means combinatory and it's given by this formula:

nCx=(n!)/((n-x)! x!)

And the expected value is given by:

E(X) =np= 162*0.75= 121.5

So we expect to win between 121 and 122 wins (121.5) at the season

Explanation:

For this case we can calculate the proportion of wins for April with the following formula:

p =(X)/(n)

Where:

X=18 number of wins in april

n = 18+6 = 24 total of games for April

And for this case we have:

p =(18)/(24)=0.75

And the proportion of losses for this month would be by the complement rule:

q = 1-p =1-0.75 =0.25

And for this case we assume that the total games for the season are 162

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest "number of games win at the season", on this case we now that: