Answer:
[tex] P(X=3)[/tex]
And we can use the probability mass function and we got:
[tex]P(X=3)=(3C3)(0.45)^3 (1-0.45)^{3-3}=0.091[/tex]
Then the probability that all three have a smart phone is 0.091
Step-by-step explanation:
Let X the random variable of interest "number of students with smartphone", on this case we now that:
[tex]X \sim Binom(n=3, p=0452)[/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 to find this probability:
[tex] P(X=3)[/tex]
And we can use the probability mass function and we got:
[tex]P(X=3)=(3C3)(0.45)^3 (1-0.45)^{3-3}=0.091[/tex]
Then the probability that all three have a smart phone is 0.091
