@blood-on-the-tracks saidDamn.
Well, the ways of picking r things from N is given by N!/(N-r)!r!
Right formula but on RHS reduced "N" to (N-1) thinking number of flavours was reduced by one too.
Nice little problem.
Well done BOT
@blood-on-the-tracks saidYes .
@venda
No problem. Have missed your weekly 'puzzle', though you did say the quality dwindled!
They've mostly been next number in the sequece type things.
I look every week and occasionally there's an interesting one.
This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?
@venda saidHmm. Probability is not my strong area.
This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?
There is an 8% chance I will figure this one out. 🙂
@venda saidI expect, that it won't die out statistically:
This weeks puzzle is an interesting one for a change!!
It concerns a lone Amoeba!!
On each hour there is an 8% probability it will die,with 56%probability it will split into two identical copies living independently and with 36% probability it will remain the same.
What are the chances that any resulting colony will eventually die out if left to it's own devices?
after one hour we have:
8 cases dead Amoeba
36% one Amoeba
56 % two Amoeba
After two hour we have:
8 + 3 11 cases of no Amoeba
13 (36% of 36) + 9 =22 cases of one amoeba (total amount of two Amobea case times 8% not taking into account that there is a non-Zero probablity for two Amoeba dying in the same time)
21 (36% of65) + 21 (56% of 36)= 42 cases of two Amoba
25 cases of three and more Amoeba.
Since only in cases of one Amoeba is a reasonable Chance of dying out and the other populations will grow I don't Thing that the case of Zero Amoeba will reach 50
@ponderable saidThe answer is quoted as a "probability" i.e 1:x
I expect, that it won't die out statistically:
after one hour we have:
8 cases dead Amoeba
36% one Amoeba
56 % two Amoeba
After two hour we have:
8 + 3 11 cases of no Amoeba
13 (36% of 36) + 9 =22 cases of one amoeba (total amount of two Amobea case times 8% not taking into account that there is a non-Zero probablity for two Amoeba dying in the sam ...[text shortened]... ing out and the other populations will grow I don't Thing that the case of Zero Amoeba will reach 50
The explanation is quite involved.
I'll leave it a day or two and then copy the explanation from the paper
@venda saidThis is the given answer:-
The answer is quoted as a "probability" i.e 1:x
The explanation is quite involved.
I'll leave it a day or two and then copy the explanation from the paper
1/7 Any new amoeba formed will have the same probability of it's colony dying out so if this probability is "p" then by considering the position at hour one,p=0.08 +0.36p +0.56p squared.From this equation p could be 1/7 or 1 but the answer 1 can be excluded as the expected number on amoebas increases each hour.