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

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

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) = (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

Answer 2

P(X>4)= 0.624

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