Question

Binomial distribution with the expression (q + p)^n where n = 5 and p = 0.3 Can you please explain how to do this?

Answer 1

@kirbykirby @nincompoop Help please?

Answer 2

Um are you trying to expand this using the binomial theorem? I'm not sure what you need since you mention the binomial distribution.

Answer 3

All it say is to use the binomial expression to calculate the binomial distribution and I don't know how to do so..

Answer 4

Do you have the full question? Do they just want you to find the probability mass function (pmf)? The binomial distribution is:
\[ f_X(x)={n\choose x}p^xq^{n-x}\] where \(q=1-p\)

If you're given p and n, then you know q... and plug in those values... then the probability of an event X=x is given by the pmf above.

Answer 5

The full question is:

Use the binomial expression (q + p)^n to calculate a binomial distribution with n = 5 and p = .3

Answer 6

Ok then I think that's what you need then. So you can calculate each probability, then you plug in values for \(x=0,1,\ldots,5\). (Notice that \(q=1-p=1-0.3 =0.7\))
So for example:

at \(x=0:\)
\[f_X(0)={5\choose 0}0.3^{5}0.7^{5-0}\]

at \(x=1:\)
\[f_X(1)={5\choose 1}0.3^{5}0.7^{5-1}\]
.... until \(x=5\)

Just need to calculate those values on your calculator

Answer 7

The confusing part of the question is that I don't know why they give you a binomial (p+q)^n. I don't see why that is needed to compute a binomial \(\textit{distribution}\), especially since that binomial expression just evaluates to 1...

Answer 8

x = 1 1,221.205 ??

Answer 9

No, each value will be between 0 and 1.

Answer 10

Oh. Okay, will you look at this?

https://answers.yahoo.com/question/index?qid=20140325093333AAzGSpG

Is this what I have to do?

Answer 11

The 2nd answer is basically what I described, but they are using the notation 5C1 instead of \(\large {5 \choose 1}\).

Answer 12

The 1st answer they just did a binomial expansion, but this is not finding a binomial distribution.

Answer 13

Okay, can yo take me through it step by step?

Answer 14

Um.. I don't know what to say other than you just need to plug in the given values, \(n\) and \(p\) into the binomial distribution distribution:
\[f_X(x)={n\choose x}p^xq^{n-x}\]
And of course, \(q=1-p\), since \(p\) is the probability of success, and we know that the sum of all probability events sum to 1. You have either a success, or a failure (whose probability is represented by \(q\)), so \(p+q=1\)

Maybe you're unsure how to calculate the binomial coefficient? That is the \(\large {n \choose x}\) term?

Answer 15

I am unsure of how to calculate that term..

Answer 16

Can you also explain the 5-0 that is the power of 7 ?

Answer 17

Do i just do 5, 4, 3, 2, 1 for x = 1 and x = 2 all the way to 5?

Answer 18

\[ {n \choose x} =\frac{n!}{x!(n-x)!}\]

So say you have \(n=8, x=3\):
\[ {8\choose 3}=\frac{8!}{3!(8-3)!}=\frac{8!}{3!\cdot 5!}\]

Here, \(8!=8\times 7 \times 6 \times 5 \times 4 \times 3\times 2 \times 1\)
Similarly,
\(5!=5 \times 4 \times 3\times 2 \times 1\)
\(3!=3 \times 2\times 1\)

So:
\[\frac{8!}{3!\cdot 5!}=\frac{8\times 7 \times 6 \times \cancel{5 \times 4 \times 3\times 2 \times 1}}{(3 \times 2\times 1)\cdot\cancel{(5 \times 4 \times 3\times 2 \times 1)}}=\frac{8\times 7 \times 6}{3\times 2}=56\]

Answer 19

Oh, okay. What about my other two question?

Answer 20

x = 0 for the first one?

Answer 21

\[f_X(0)={5\choose 0}0.3^{5}0.7^{5-0}\]... Here I just substituted the \(x\) with 0.
Be careful, here you have the decimal numbers 0.3 and 0.7 raised to a power. To make it more clear:
\[f_X(0)={5\choose 0}(0.3)^{0}(0.7)^{5-0}\]
I should point out at this stage that ... I made a typo when I inputted "5" in the exponent of 0.3 Sorry about that =( There is clearly an \(x\) there in the formula.

I just left the 5-0 there, rather than just 5, just to show you what you need to do in the formula.

Now for x=1: just do:
\[f_X(1)={5\choose 1}(0.3)^{1}(0.7)^{5-1}=5(0.3)(0.7)^4\]

Now for x=2: just do:
\[f_X(2)={5\choose 2}(0.3)^{2}(0.7)^{5-2}=10(0.3)^2(0.7)^3\]

Answer 22

Okay, is this right?

Answer 23

yes looks good, although in the first line you wrote \(\large {5 \choose 0}\) even though you still correctly put 1 instead of 0 in subsequent steps ;)

Answer 24

Okay. Ha thanks, dang so now I have to do all that 4 more times?

Answer 25

Yeah. but also include the case for x=0.

Answer 26

Alright well thank you very much! Just one more thing, at the very end after I do all the rest do I need to write 1 through 5 in just answer form?

Could you hope me with one more?

Answer 27

*help

Answer 28

I suppose you could? I'm not sure how detailed your instructor needs your answers to be :S

Answer 29

Sure

Answer 30

Ha okay. I'll open it in another post so you get both medals :P

Answer 31

Hehe ok