Parmi | #1 draft pick | Jake King is **** | Big Bash League tipping champion of the universeCome and Paint Turtle
LOL, you're right.
Greatest Ever Test XI: JB Hobbs, L Hutton, DG Bradman (c), IVA Richards, BC Lara, GS Sobers, AC Gilchrist (wk), Imran Khan, RJ Hadlee, MD Marshall, SK Warne 12th man: M Muralitharan
Favorite XI: WG Grace, VT Trumper, IVA Richards, DCS Compton, FMM Worrell (c), AC Gilchrist (wk), CL Cairns, SK Warne, FS Trueman, SE Bond, T Richardson 12th man: H Larwood
"Neither of them will have an international cricket acareer past 2016."
Brocky on Martin Guptill and Ish Sodhi. 20/11/2014.
"I want to raise my hand and say one thing. Those who complain about my love for the game or commitment to the game are clueless. These are the only 2 areas where I give myself 100 out of 100."
- Sachin Tendulkar, as told in an interview published in Bengali newspaper Anandabazar Patrika after his 100th International century (translated by weldone)
Hulululu has 3/4th chance of dying and 1/4th chance of breaking into 2 Hulululus. If broken into 2, each of these 2 Hulululus have the same chances - 3/4th chance of dying and 1/4th chance of breaking into 2 further Hulululus....and so on.
What's the probability that the Hulululu species will never be extinct?
I'm massively sleep deprived but: 0. Monkeys, Shakespeare, etc.
Yeah the sum you'd set up would converge to 1, wouldn't it?
**** I hate probability. Prefer something nice and clean like finding two irrationals a and b such that a^b is rational.
+ time's fickle card game ~ with you and i +
get ready for a broken ****in' arm
My immediate reaction was to keep it simple and go with tree diagrams, but the splitting in two part threw me off.
Immediately I thought it was (1/4)^n with n becoming very large, but obviously that doesn't work.
Secondarily I thought in terms of series; the probability being quartered each step as there are two, but I doubt that holds up mathematically. That would have been 1/4 * 1/16 * 1/64 ...
So whatever the formula is [I got P(n)=1/(4^n), but I may be very wrong], as n becomes very large you approach zero, so it is almost certain they will become extinct at some point.
Last edited by Dan; 27-12-2012 at 07:34 AM.
Repeating patterns, mate.
OK. Assume 'if we start with a bacteria, the probability of the species finally going extinct is p'. Then,
p = 3/4 + 1/4*p*p
=> (p-2)^2 = 1
=> p = 1 [Rejecting the other solution as that can't be a probability number]
So, the probability of of the Hulululu species never going extinct = 1-p = 0
Last edited by weldone; 27-12-2012 at 08:00 AM.
In fact, for any x where 0<x<1, if you replace 3/4 by x and 1/4 by (1-x) in the original problem, then also p will come out as 1 from the equation I think. I'm also sleep-deprived, so not putting an effort to solve that generic one lol.
Alternatively you can look at the expected population at every year.
E(0) = 1
E(1) = 0.5
E(2) = 0.25
E(3) = 0.125
MSN Messenger: minardineil2000 at hotmail dot com | AAAS Chairman
CricketWeb Black | CricketWeb XI Captain
ClarkeWatch: We're Watching Rikki - Are You?
Up The Grecians - Exeter City FC
Completing the Square: My Cricket Web Blog
Also love the fact that he wasn't completely correct.
There are currently 1 users browsing this thread. (0 members and 1 guests)