What Is the Minimum Number of Articles Needed for a 75% Acceptance Probability?

[luv2learn]

Homework Statement

The rejection rate of a certain journal is 45%. If the journal accepts articles at random, what is the minimum number of articles someone has to submit to have a probability of more than 0.75 of getting at least one article accepted?

Homework Equations

I'm almost sure this is a binomial distribution question where you take p and n to kook up the P(X) in the binomial probabilities table. Only thing is, I don't know what is n.

The Attempt at a Solution

p=1-0.45=0.55

P(1) = 1-P(X<=0)
>0.75 = 1-P(X<=0)
P(X<=0) < 0.25

But then what? Is my potential n the minimum nr of articles or 1?

{{Also, this is my first post, would someone please tell me where to get the scientific notation for the formulas to put in the posts? Pls and tx! }}}

 

[tiny-tim]

Welcome to PF!

Hi luv2learn! Welcome to PF!

(have an leq: ≤ )

No, it's not binomial …

you're right (if I'm reading you properly: your notation is a bit weird ) that the question is the same as what is 1 - Q_n,

where Q_n is the probability that all n articles are rejected.

ok, rejections are independent, so what is Q_n ? ​

 

[luv2learn]

Ok, so apparently I've got this whole question wrong, LOL

So the probability that n artiles are rejected is Q_n = 0.45 x n

 

[tiny-tim]

So the probability that n artiles are rejected is Q_n = 0.45 x n

erm … with n = 3, that's greater than 1 !

Try again! ​

 

[luv2learn]

(I'm really losing it, been at it for 10hrs.)

Q_n=0.45ⁿ ;
Rejecting 1 is: Q₁=0.45¹; Which implies accepting n-1, which is = 1-0.45¹ = 0.55
Q₂=0.45²; accept n-2 = 1-0.45² = 0.798; etc.

So if x = minimum nr of articles to be submitted, then I'm actually trying to find
Accept n-x = 1-0.45^x > 0.75 ?

 

[tiny-tim]

Now you're confusing me …

you're looking for n such that 0.45ⁿ < 0.25

(either use logs or just trial-and-error! )

 

[luv2learn]

Yeah, tx. I got the same thing but in a very long (and confusing) way.
In the end n > 1.74 i.e. n = 2

Tx a lot. But is there a simple way of seeing if its a binomial distribution or not? I thought I know but clearly I don't. Or can the same answer be reached if I use binomial distribution probability rules?

 

[tiny-tim]

(How did you get 1.74? )

You're misunderstanding which bit of the binomial is which.

For (p + q)ⁿ, the figure for k successes is p^kq^{n-k n}C_k …

in this case, technically, you did use the binomial theorem, but with k = n and therefore ⁿC_k = 1.

 

[luv2learn]

0.45ⁿ>0.25
log (0.45ⁿ>log (0.25)
nlog(0.45)>log(0.25)
n=log0.25/log0.45
n=1.736

 

[tiny-tim]

oh yes, that's fine.