Question

The Magazine Mass Marketing Company has received 1515 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.50.5. What is the probability that less than a third of the entry forms will include an order? Round your answer to four decimal places.

Answer 1

Answer:

$P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)$

$P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305$

$P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458$

$P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032$

$P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139$

$P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417$

And adding we got:

$P(x < 5)=0.0592$

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:

$X \sim Binom(n=15, p=0.5)$

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 to find this probability:

$P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)$

$P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305$

$P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458$

$P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032$

$P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139$

$P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417$

And adding we got:

$P(x < 5)=0.0592$