Answer :
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
[tex]P(X=x)=(_{x}^{n})\ p^xq^{n-x}[/tex]
[tex]P(X=0)=(_{0}^{10})\ 0.5^0\ 0.5^{10}=0.0009[/tex]
[tex]P(X=1)=(_{1}^{10})\ 0.5^1\ 0.5^{9}=0.0097[/tex]
[tex]P(X=2)=(_{2}^{10})\ 0.5^2\ 0.5^{8}=0.043[/tex]
[tex]P(X=3)=(_{3}^{10})\ 0.5^3\ 0.5^{7}=0.117[/tex]
[tex]P(X=4)=(_{4}^{10})\ 0.5^3\ 0.5^{7}=0.205[/tex]
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:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]p = 0.5[/tex]
What is the probability that more than two-fifths of the entry forms will include an order?
Two-fifths of 10 is 4. So
[tex]P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{10,5}.(0.5)^{5}.(0.5)^{5} = 0.2461[/tex]
[tex]P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051[/tex]
[tex]P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172[/tex]
[tex]P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439[/tex]
[tex]P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098[/tex]
[tex]P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.001[/tex]
[tex]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[/tex]
0.6231 = 62.31% probability that more than two-fifths of the entry forms will include an order