Respuesta :
Answer:
a) [tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332[/tex]
b) [tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649[/tex]
[tex]P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099[/tex]
[tex]P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040[/tex]
[tex]P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465[/tex]
[tex]P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732[/tex]
[tex]P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228[/tex]
[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033[/tex]
And replacing we got:
[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175 [/tex]
c) [tex] P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)[/tex]
And replacing we got 0.825
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=17, 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]
Part a
And we want to find this probability:
[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332[/tex]
Part b fewer than 10 will check bags
We want this probability:
[tex] P(X<10) [/tex]
We can use the complement rule and we have:
[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649[/tex]
[tex]P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099[/tex]
[tex]P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040[/tex]
[tex]P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465[/tex]
[tex]P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732[/tex]
[tex]P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228[/tex]
[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033[/tex]
And replacing we got:
[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175 [/tex]
Part c at least 10 bags
We can find this probability like this:
[tex] P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)[/tex]
And replacing we got 0.825
