Question

In a list of households, own homes and do not own homes. Five households are randomly selected from these households. Find the probability that the number of households in these who own homes is exactly .

Answer 1

Answer:

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=15, p=0.6)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

And we want this probability:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$

Step-by-step explanation:

Adduming the following info: In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

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

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=15, p=0.6)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

And we want this probability:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$