Question

Use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3.

Answer 1

well the formula I use is basically

\[^nC_{r} \times p^x \times (1 - p)^{n - r}\]

so you know n and p but not r

Answer 2

How do i set this up? This is for extra credit, and i really don't understand.

Answer 3

binomial expansion

Answer 4

Yes

Answer 5

@campbell_st why cant she use q=1-p and cancel everything and say its just 1^5=1
because the odds of P+Q which are complimentary cases happening should that be one anyway

Answer 6

you can use q... its just for some people finding q or understanding what it is... can be difficult... so I use (1 - p)

they have the same value...

Answer 7

This is like latin to me you guys.

Answer 8

well based on the info provided

its

\[^5C_{r} (0.3)^r(1 - 0.3)^{5 - r}\]

Answer 9

do what campbell says you can trust him

Answer 10

The equation in my book is super long and doesn't look like that…

Answer 11

yes queen you have to include all the terms

Answer 12

I literally wrote the entire question. Thats all the info they gave me.

Answer 13

5C0 p^5 + 5C1 p^4 q +5C2 p^3 q^2 +5C3 p^2 q^3 +5C4 p^1 q^4 +5C5 q^5

Answer 14

so for the expansion yon need to let

r = 0, 1, 2, 3, 4, 5

so there are 6 terms

\[(0.3)^5 + 5 \times (0.3)^4\times(0.7)+ 10 \times (0.3)^3\times (0.7)^2 + 10 \times (0.3)^2 \times (0.7)^3 + 5 \times (0.3) \times (0.7)^4 + (0.7)^5\]

Answer 15

Thats what it looks like!!! okay!

Answer 16

thats should equal to 1, if im not mistaken let me go check wolfram

Answer 17

the missing bit is

\[+ 5 \times (0.3) \times(0.7)^4 + (0.7)^5\]

Answer 18

http://www.wolframalpha.com/input/?i=%280.3%29%5E5+%2B+5+%5Ctimes+%280.3%29%5E4%5Ctimes%280.7%29%2B+10+%5Ctimes+%280.3%29%5E3%5Ctimes+%280.7%29%5E2+%2B+10+%5Ctimes+%280.3%29%5E2+%5Ctimes+%280.7%29%5E3+%2B+5+%5Ctimes+%280.3%29+%5Ctimes+%280.7%29%5E4+%2B+%280.7%29%5E5

Answer 19

its 1 which is right!

Answer 20

So my answer is just 1?

Answer 21

i think they were just trying to make you to think that
since its (P+q)^n = (p+(1-p)^n = 1^n so you can see it must be 1 no matter what

Answer 22

Oh.. okay. Well thanks so much!

Answer 23

ok sure, yw!

Answer 24

Do you have the steps you took to get this answer? I have to type a paragraph (for math i know how dumb) on this and i need to explain that. Thanks Doll.

Answer 25

@dan815

Answer 26

its a long explaination if you havent seen pascals triangle of this n Choose k stuff

Answer 27

or*

Answer 28

oh wait scratch that okay

Answer 29

the whole point is this binomial distribution, we get to know how much of each case is taking place

Answer 30

Okay..

Answer 31

like campbell wrote that formula... with 5c1, 5c2, 5c3...
each one of those is telling you how this thing is distributed, like the prob of winning twice and failing 3 times or winning 3 times and failing twice and so on

Answer 32

https://www.google.ca/search?q=binomial+distribution+of+p+%2B+q&source=lnms&tbm=isch&sa=X&ei=ZkSNU83tNMmj8QG7uYHYDw&sqi=2&ved=0CAYQ_AUoAQ&biw=1093&bih=515#facrc=_&imgdii=_&imgrc=9KRTBMAlT8ApRM%253A%3Byf8Cw6x8I_ZFtM%3Bhttp%253A%252F%252Fwww.windows2universe.org%252Fspaceweather%252Fimages%252Frandom_walk_graph.gif%3Bhttp%253A%252F%252Fwww.windows2universe.org%252Fspaceweather%252Faoss335_quietsun5a.html%3B576%3B469

Answer 33

if u plot the terms separating the cases you will see this kind of a graph where in your case you will see 5 lines since n = 5