If a car agency sells 50% of its inventory of a certain foreign car equipped with side airbags, find a formula for the probability distribution of the number of cars with side airbags among the next 4 cars sold by the agency.

Respuesta :

Answer:

Let X the random variable of interest "number of cars with side airbags among the next 4", on this case we now that:

[tex]X \sim Binom(n=4, p=0.5)[/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 the distribution is given by:

[tex] P(X)= (4CX) (0.5)^x (1-0.5)^{4-x}[/tex]

With [tex] X=0,1,2,3,4[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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 "number of cars with side airbags among the next 4", on this case we now that:

[tex]X \sim Binom(n=4, p=0.5)[/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 the distribution is given by:

[tex] P(X)= (4CX) (0.5)^x (1-0.5)^{4-x}[/tex]

With [tex] X=0,1,2,3,4[/tex]

Answer:

f(x) = ⁴Cₓ (0.0625) (for 0 ≤ x ≤ 4)

Step-by-step explanation:

This is a binomial distribution problem

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 4

x = Number of successes required = x

p = probability of success = probability of selling a car with airbag = 0.5

q = probability of failure = probability of selling a car without airbag = 1 - 0.5 = 0.5

So, for any x,

P(X=x) = ⁴Cₓ (0.5)ˣ (0.5)⁴⁻ˣ

On simplification,

P(X=x) = ⁴Cₓ (0.5)⁴⁻ˣ⁺ˣ = ⁴Cₓ (0.5)⁴

= ⁴Cₓ (0.0625)

So, the frequency distribution for number of cars with side airbags among the next 4 cars sold by the agency is given by

f(x) = ⁴Cₓ (0.0625) (for 0 ≤ x ≤ 4)

Hope this Helps!!!