Question

Give the numerical value of the parameter p in the following binomial distribution scenarioA softball pitcher has a 0.721 probability of throwing a strike for each pitch and a 0.279 probability of throwing a ball. If the softball pitcher throws 19 pitches, we want to know the probability that more than 15 of them are strikes.

Answer 1

Answer:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19)$

$P(X=16)=(19C16)(0.721)^{16} (1-0.721)^{19-16}=0.112$  

$P(X=17)=(19C17)(0.721)^{17} (1-0.721)^{19-17}=0.051$  

$P(X=18)=(19C18)(0.721)^{18} (1-0.721)^{19-18}=0.015$  

$P(X=19)=(19C19)(0.721)^{19} (1-0.721)^{19-19}=0.002$  

And replacing we got:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19) =0.112+0.051+0.015+0.002= 0.1801$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

For this case our random variable is given by:

$X \sim Binom(n = 19, p = 0.721)$

For this case we want this probability:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19)$

$P(X=16)=(19C16)(0.721)^{16} (1-0.721)^{19-16}=0.112$  

$P(X=17)=(19C17)(0.721)^{17} (1-0.721)^{19-17}=0.051$  

$P(X=18)=(19C18)(0.721)^{18} (1-0.721)^{19-18}=0.015$  

$P(X=19)=(19C19)(0.721)^{19} (1-0.721)^{19-19}=0.002$  

And replacing we got:

$P(X>15)= P(X=16)+P(X=17)+P(X=18)+P(X=19) =0.112+0.051+0.015+0.002= 0.1801$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$