iam looking for optimzed naval combat strategies atm
unfortunately i realized that my ten years of math in school ages ago are not enough and iam rusty as hell
I need to calculate the probability of a event (not) to happen. I think its acutally very easy but i never learned it and so i have no real idea how to do it.
Could someone point me in the right direction please?
For event A to happen, 1 out of z attemps needs to be successfull.
Statistical speaking, 1 in 3 attemps is successfull.
How likely is it that event A will not happen, if you try X z times?
if i draw up every possible case on paper and add everything up, i get a probability of event A occurring of
0,33 for z = 1 (obviously since one in three attempts will result in a kill )
0,56 for z = 2
0,7 for z = 3
0,81 for z = 4
Or in Defconian:
If a bomber is being shot at (with a fighter) four times, the bomber will be killed in 81 of 100 attemps.
The probability of surviving is below 20%.
Is that right?
And if yes, how does it look if you shoot at a bomber 12 times?
or to be precise, how does it look like if 12 fighters shoot at the same bombers once simultaneously?
How can i calculate that one?
*cries for help*
simple math problem
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- Nightwatch
- level5
- Posts: 1288
- Joined: Wed Sep 19, 2007 3:02 pm
- Location: Germany
The probability of an even happening in one attempt is x (in the format 1=surely will happen, 0=surely will not happen).
Then the probability of it NOT HAPPENING is 1-x.
The probability of it NOT HAPPENING in y attempts is: (1-x)^y.
Example: I'm throwing a coin 5 times. Let's calculate the probability of side "A" NOT happening 5 times in a row:
x=0,5
y=5
Probability= (1-0,5)^5 = 1 / (2^5) = 1/32.
Same example with a fighter-bomber case: x=0,(6) (survives 2 out of 3 attempts), y=4 (4 attempts)
then the probablility of it surviving 4 shots is (2/3)^4=16/81=0,1975
For 12 attempts, it looks bad for a bomber
(2/3)^12=4096/531441=0,00771
Then the probability of it NOT HAPPENING is 1-x.
The probability of it NOT HAPPENING in y attempts is: (1-x)^y.
Example: I'm throwing a coin 5 times. Let's calculate the probability of side "A" NOT happening 5 times in a row:
x=0,5
y=5
Probability= (1-0,5)^5 = 1 / (2^5) = 1/32.
Same example with a fighter-bomber case: x=0,(6) (survives 2 out of 3 attempts), y=4 (4 attempts)
then the probablility of it surviving 4 shots is (2/3)^4=16/81=0,1975
For 12 attempts, it looks bad for a bomber
(2/3)^12=4096/531441=0,00771
- Nightwatch
- level5
- Posts: 1288
- Joined: Wed Sep 19, 2007 3:02 pm
- Location: Germany
Re: simple math problem
Hey guys! Do you aware of any sites where I can find math worksheets?
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- level2
- Posts: 127
- Joined: Wed Jun 15, 2022 11:53 pm
Re: simple math problem
Hi all! I think every child has problems with numbers, some can count to 50, some can count to 20, and some can't even count to 8. If your child has this problem, I can offer you number 8 worksheets from Brighterly online school. I am sure that they will help your child not only with numbers but also with all the other topics that they are having a hard time with.
I think I have helped you with your question.
I think I have helped you with your question.
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