Question

The Magazine Mass Marketing Company has received 14 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.4. What is the probability that no less than 4 of the entry forms will include an order?

Answer 1

0.8759 is the probability that no less than 4 of the entry forms will include an order.

Explanation:

We are given the following information:

We treat subscription order with an entry form as a success.

P(subscription order with an entry form) = 0.4

Then the number subscription order with an entry form follows a binomial distribution, where

`P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)`

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 14

We have to evaluate:

`P(x \geq 4) =1 - P(x = 0) - P(x = 1) - P(x = 2) - P(x = 3) \\= 1 - (\binom{14}{0}(0.4)^0(1-0.4)^(14) + \binom{14}{1}(0.4)^1(1-0.4)^(13) + \binom{14}{2}(0.4)^2(1-0.4)^(12) + \binom{14}{3}(0.4)^3(1-0.4)^(11))\\=1-(0.0007+0.0074+0.0317+0.0846)\\= 0.8756`

0.8759 is the probability that no less than 4 of the entry forms will include an order.