Question

There is a 70 percent chance that an airline passenger will check bags. In the next 16 passengers that check in for their flight at Denver International Airport (a) Find the probability that all will check bags. (Round your answer to 4 decimal places.) P(X

Answer 1

Answer:

a) $P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

b) $P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)]$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175$

c) $P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

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:

$X \sim Binom(n=17, p=0.7)$

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:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

And we want to find this probability:

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

Part b fewer than 10 will check bags

We want this probability:

$P(X<10)$

We can use the complement rule and we have:

$P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)]$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175$

Part c at least 10 bags

We can find this probability like this:

$P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825