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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
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](https://tex.z-dn.net/?f=p%20%3D%200.5)
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)](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29%20%2B%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 5) = C_{10,5}.(0.5)^{5}.(0.5)^{5} = 0.2461](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20C_%7B10%2C5%7D.%280.5%29%5E%7B5%7D.%280.5%29%5E%7B5%7D%20%3D%200.2461)
![P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051](https://tex.z-dn.net/?f=P%28X%20%3D%206%29%20%3D%20C_%7B10%2C6%7D.%280.5%29%5E%7B6%7D.%280.5%29%5E%7B4%7D%20%3D%200.2051)
![P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172](https://tex.z-dn.net/?f=P%28X%20%3D%207%29%20%3D%20C_%7B10%2C7%7D.%280.5%29%5E%7B7%7D.%280.5%29%5E%7B3%7D%20%3D%200.1172)
![P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.5%29%5E%7B8%7D.%280.5%29%5E%7B2%7D%20%3D%200.0439)
![P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.5%29%5E%7B9%7D.%280.5%29%5E%7B1%7D%20%3D%200.0098)
![P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.001](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.5%29%5E%7B10%7D.%280.5%29%5E%7B0%7D%20%3D%200.001)
![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](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29%20%2B%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.2461%20%2B%200.2051%20%2B%200.1172%20%2B%200.0439%20%2B%200.0098%20%2B%200.0010%20%3D%200.6231)
0.6231 = 62.31% probability that more than two-fifths of the entry forms will include an order