Sy123
This too shall pass
- Joined
- Nov 6, 2011
- Messages
- 3,730
- Gender
- Male
- HSC
- 2013
Last edited:
How did you get (2r-1)/(r^3-r), that is the correct answer though, I split the 1/k(k+2) to 1/2 ( 1/k - 1/(k+2)), then manipulated and grouped certain terms to telescope twice.Part i: first step change it to 11/8 + Sum from r=3 to infinite of (2r-1)/(r^3-r)
Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4 as the answer.
Intermediate steps included:
11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....)
Which equals
11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r)
Sorry no Latex - still working on my skills in that department.
I don't think soThis one's the triangle inequality right? Just need a little bit of induction to extend it.
Ohhh I see what you mean now, nice work.
Here is an attempt:There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any.
i) What is the probability of me drawing out K red marbles?
ii) What is the most likely number of red marbles I will draw out?
iii) What is the average number of red marbles I will draw out?
i) is correct.Here is an attempt:
Ahh ok, I'm not too sure about how to interpret 'average' though..i) is correct.
ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1
Ahh ok, I'm not too sure about how to interpret 'average' though..
If I did:
And took the average probability:
And then, if:
(so its bounded by two consecutive probabilities)
Then the one its closest to, (say its closest to P(m)), that means m is the average number of balls????
Oh ok