Question

The Magazine Mass Marketing Company has received 10 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.5. What is the probability that more than two-fifths of the entry forms will include an order? Round your answer to four decimal places.

Answer 1

Answer:

P(X>4)= 0.624

Step-by-step explanation:

Given that

n = 10

p= 0.5 ,q= 1 - p = 0.5

Two fifth of 10 = 2/5 x 10 =4

It means that we have to find probability P(X>4).

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

We know that

$P(X=x)=(_{x}^{n})\ p^xq^{n-x}$

$P(X=0)=(_{0}^{10})\ 0.5^0\ 0.5^{10}=0.0009$

$P(X=1)=(_{1}^{10})\ 0.5^1\ 0.5^{9}=0.0097$

$P(X=2)=(_{2}^{10})\ 0.5^2\ 0.5^{8}=0.043$

$P(X=3)=(_{3}^{10})\ 0.5^3\ 0.5^{7}=0.117$  

$P(X=4)=(_{4}^{10})\ 0.5^3\ 0.5^{7}=0.205$

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205

P(X>4)= 0.624

Answer 2

Answer:

0.6231 = 62.31% probability that more than two-fifths of the entry forms will include an order

Step-by-step explanation:

For each entry, there are only two possible outcomes. Either there is a subscription order, or there is not. The probability of an entry having a subscription order is independent of other entries. So we use the binomial probability distribution to solve this quesiton.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

10 entries in its latest sweepstakes.

This means that n = 10.

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that $p = 0.5$

What is the probability that more than two-fifths of the entry forms will include an order?

Two-fifths of 10 is 4. So

$P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.5)^{5}.(0.5)^{5} = 0.2461$

$P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051$

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

$P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439$

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.001$

$P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2461 + 0.2051 + 0.1172 + 0.0439 + 0.0098 + 0.0010 = 0.6231$

0.6231 = 62.31% probability that more than two-fifths of the entry forms will include an order