Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

p=.45
q=.55

I guess I have to find the minimum n

I think I got this...one moment plz

I meant P[0]+P[1]+P[3]

shouldn't it be p(0),p(1) and p(2)

90% sure of filling the quota, so isn't p=0.9?

What does P(r) represent here?

so n=6 and u need to find r ?

no that was for part (a) of the problem... this is part (e)

So you wouldn't use the whole formula?
\[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]?

sorry r>=3

in other words solve for n?

yup.

Great! That makes sense. I'll do that calculation and get back to ya...just a min...

ok.

hmmm....doesn't look pretty....
\[3!\frac{0.9}{0.45}^3=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

u don't get 0.9^3

oops typo
\[3!\frac{0.9}{0.45^3}=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

yup, 0.55^n is not pretty at all
n!=n(n-1)(n-2)(n-3)!

This looks like a negative binomial situation.
b*(x; r, P) = x-1Cr-1 * Pr * (1 - P)x - r

uhm. here is what I found online....
"Since the probability of success is p=.45, we need to look in the binomial table under p=0.45 and different values of n that will satisfy this preceding relation. Binomial Probability Distribution table shows that if n=10 when p=0.45, then
\[P(r\ge3)=1-P(r<3)\]
\[=1-(P(0)+P(1)+P(2))\]
\[=1-(0.003+0.021+0.076)\]
\[=1-0.10\]
\[=0.90\]

yes, they used a table, then verified the result

Doesn't your text give a hint as to how to solve this type of problem?