Respuesta :
Answer:
We can begin calculating the following probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got:
[tex] P(X=3)= 3C3 (0.7)^3 (1-0.7)^{3-3}= 0.343[/tex]
And since we want to find the probability of not getting all strikes in three attempts we can ose the complement rule and we got:
P(not getting all strikes in three attempts) =1-P(X=3) =1-0.343=0.657
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=3, p=0.7)[/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]
We can begin calculating the following probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got:
[tex] P(X=3)= 3C3 (0.7)^3 (1-0.7)^{3-3}= 0.343[/tex]
And since we want to find the probability of not getting all strikes in three attempts we can ose the complement rule and we got:
P(not getting all strikes in three attempts) =1-P(X=3) =1-0.343=0.657
