Question

The Magazine Mass Marketing Company has received 12 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.6. What is the probability that more than a fourth of the entry forms will include an order? Round your answer to four decimal places.

Answer 1

Answer:

$P(X>3) = 1-P(X\leq 3) =1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$  

And we can find the individual probabilities like this:

$P(X=0)=(12C0)(0.6)^0 (1-0.6)^{12-0}=0.0000168$  

$P(X=1)=(12C1)(0.6)^1 (1-0.6)^{12-1}=0.000302$  

$P(X=2)=(12C2)(0.6)^2 (1-0.6)^{12-2}=0.00249$  

$P(X=3)=(12C3)(0.6)^3 (1-0.6)^{12-3}=0.01246$  

And replacing we got:

$P(X>3) =1-[0.0000168+0.000302+0.00249+0.01246] =0.9847$

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:  

$X \sim Binom(n=12, p=0.6)$  

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:  

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

For this case we want this probability:

$P(X> 3)$

Since one fourth of the total number 12 is 12/4 =3

We can find this probability using the complement rule:

$P(X>3) = 1-P(X\leq 3) =1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$  

And we can find the individual probabilities like this:

$P(X=0)=(12C0)(0.6)^0 (1-0.6)^{12-0}=0.0000168$  

$P(X=1)=(12C1)(0.6)^1 (1-0.6)^{12-1}=0.000302$  

$P(X=2)=(12C2)(0.6)^2 (1-0.6)^{12-2}=0.00249$  

$P(X=3)=(12C3)(0.6)^3 (1-0.6)^{12-3}=0.01246$  

And replacing we got:

$P(X>3) =1-[0.0000168+0.000302+0.00249+0.01246] =0.9847$

Answer 2

Answer:

Probability that more than one fourth of the entry forms will include an order is 0.9847.

Step-by-step explanation:

We are given that the Magazine Mass Marketing Company has received 12 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.6.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 12 entries

            r = number of success = more than one fourth which is equal to $\frac{12}{4}$ =

                                                     3 entry forms

  p = probability of success which in our question is probability of

       receiving a magazine subscription order with an entry form, i.e; 0.60

LET X = Number of entry forms that will include an order

So, it means X ~ $Binom(n=12, p=0.60)$

Now, Probability that more than one fourth of the entry forms will include an order is given by = P(X > 3)

 P(X > 3)  = 1 - P(X $\leq$ 3)

               = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= $1- [\binom{12}{0}\times 0.60^{0} \times (1-0.60)^{12-0} + \binom{12}{1}\times 0.60^{1} \times (1-0.60)^{12-1} +\binom{12}{2}\times 0.60^{2} \times (1-0.60)^{12-2}+\binom{12}{3}\times 0.60^{3} \times (1-0.60)^{12-3}]$

= $1-[ 1 \times 1 \times 0.40^{12}+12 \times 0.60^{1} \times 0.40^{11}+66 \times 0.60^{2} \times 0.40^{10} +220 \times 0.60^{3} \times 0.40^{9} ]$

= 1 - 0.01527 = 0.9847

Therefore, Probability that more than a fourth of the entry forms will include an order is 0.9847.