Respuesta :
Answer:
[tex] P(X=3)[/tex]
And we can use the probability mas function and we got:
[tex]P(X=3)=(9C3)(0.51)^3 (1-0.51)^{9-3}=0.1542[/tex]
Step-by-step explanation:
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, on this case we now that:
[tex]X \sim Binom(n=9, p=0.51)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want this probability:
[tex] P(X=3)[/tex]
And we can use the probability mas function and we got:
[tex]P(X=3)=(9C3)(0.51)^3 (1-0.51)^{9-3}=0.1542[/tex]
Answer:
15.42% probability that she gets exactly 3 successful first serves in
Step-by-step explanation:
For each first serve there are only two possible outcomes. Either it is successful, or it is not. The probability of a first serve being successful is independent of other first serves. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A tennis player makes a successful first serve 51% of the time
This means that [tex]p = 0.51[/tex]
If she serves 9 times, what is the probability that she gets exactly 3 successful first serves in?
This is [tex]P(X = 3)[/tex] when [tex]n = 9[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{9,3}.(0.51)^{3}.(0.49)^{6} = 0.1542[/tex]
15.42% probability that she gets exactly 3 successful first serves in
