Math Help - I dont know how to find these answers for these problems...probability

1. I dont know how to find these answers for these problems...probability

2. A process follows the binomial distribution with n = 7 and p = .4. Find
a. P(x = 3)
b. P(x > 5)
c. P(equal than or less than)2 (I dont have that keyboard symbol sorry)

2. Originally Posted by jwells1999
2. A process follows the binomial distribution with n = 7 and p = .4. Find
a. P(x = 3)
b. P(x > 5)
c. P(equal than or less than)2 (I dont have that keyboard symbol sorry)
If you are working on these types of questions, you must have been taught that for a binomial distribution with n trials and probability of success p in a single trial:

$Pr(X = x) = {^nC_x} \, p^x (1 - p)^{n-x}$

a. Apply the above formula.

b. Pr(X = 6) + Pr(X = 7). Apply the above formula on each.

c. Pr(X = 0) + Pr(X = 1) + Pr(X = 2). Apply the above formula on each.

And please do not double post.

3. First of all I just wanted to say sorry for the double post, I didnt know where to post my question.

As with the problem...

I was taught a different formula and I think thats what is getting me confused.
I was taught f(x)=n!/x!(n-x)!*p^x(1-p)^n-x

For a and c I got the following result
a)F(x)=5040/144*.064(.6)^4
F(x)=35*.064*.1296
F(x)=35*.0082944
F(x)=.290304 or 29.03%

c)F(x)=(7!) / (2!(7-2)! * (.4^2)(.6)^5
F(x)=[5040/240]*.16(.07776)
F(x)=21*.0124416
F(x)=.2612736 or 26.13%

but I cant figure out b. Should I be solving whereas x=6 since the problem is x>5??

If so I would get

F(x)=(7!) / (6!)(7-6)! *.4^6(1-.4)^7-6
F(x)=(7)*.004096(.6)
F(x)=7*.0024576 or .25%

but Im not sure if thats the right way to do it.

Any help would be much appreciated
Thanks

4. Originally Posted by jwells1999
First of all I just wanted to say sorry for the double post, I didnt know where to post my question.

As with the problem...

I was taught a different formula and I think thats what is getting me confused.
I was taught f(x)=n!/x!(n-x)!*p^x(1-p)^n-x

Mr F says: $\frac{n!}{x! (n - x)!}$ is the same as ${^nC_x}$. In fact, it's the basic formula for ${^nC_x}$

For a and c I got the following result
a)F(x)=5040/144*.064(.6)^4
F(x)=35*.064*.1296
F(x)=35*.0082944
F(x)=.290304 or 29.03%

c)F(x)=(7!) / (2!(7-2)! * (.4^2)(.6)^5
F(x)=[5040/240]*.16(.07776)
F(x)=21*.0124416
F(x)=.2612736 or 26.13%

but I cant figure out b. Should I be solving whereas x=6 since the problem is x>5?? Mr F says: read what I posted for this question.

If so I would get

F(x)=(7!) / (6!)(7-6)! *.4^6(1-.4)^7-6
F(x)=(7)*.004096(.6)
F(x)=7*.0024576 or .25%

but Im not sure if thats the right way to do it.

Any help would be much appreciated
Thanks
..