Probobility Question? MATH - Printable Version

Probobility Question? MATH - Printable Version

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Probobility Question? MATH by .Roy on 02-14-2006 at 05:59 AM

Well I just forgot this from when i learned it like 3 years ago -.- And I became curious to find out what the formula is to, well let me explain.

Lets say u have a 60% chance of getting something you want, and you get 2 tries. Whats the chance that in ONE of the tries you will get it.

RE: Probobility Question? MATH by rix on 02-14-2006 at 06:17 AM

Logically it would be 30%?! But heey, im not a rocket scientist.

RE: Probobility Question? MATH by ddunk on 02-14-2006 at 06:26 AM

(6/10)*(6/10) = 0.36

36%

RE: Probobility Question? MATH by .Roy on 02-14-2006 at 07:33 AM

Omfg, Im sorry for being mad when you guys are trying to help. But PLEASE read the fukin question.

I want it to happend once not both times. if it was both times its just the percentage times percentage like DDunk said.

But thats not the questions, so read it.

RE: Probobility Question? MATH by Plik on 02-14-2006 at 07:52 AM

Probability of it happening just once = 12/25 = 48%
Probability of it happening atleast once = 21/25 = 84%

I think thats right but its been a while since i did maths

Its easy to work out using tree diagrams

RE: Probobility Question? MATH by Chrono on 02-14-2006 at 08:18 AM

quote:
Originally posted by Plik
Probability of it happening just once = 12/25 = 48%

(i think

)

P(A)=0,6
P(B)=0,6

P(AvB) = P(A)+P(B) - 2*P(A^B) = 0,6 + 0,6 - 2*0,36 =1.2 - 0,72 = 0,48 = 48%

i think that's how you do it :-/ (that would be the probability of getting it JUST on one try (not on AT LEATS one try), it's funny how you can know really advanced maths and forget about such simple stuff :P

RE: Probobility Question? MATH by RaceProUK on 02-14-2006 at 10:55 AM

P (A) = 0.6
P (B) = 0.6

To get the correct probaboility, you need:
P (A) +P(¬A)*P (B)
= 0.6 + 0.4*0.6
= 0.6 + 0.24
= 0.84 = 84%

RE: Probobility Question? MATH by Mnjul on 02-14-2006 at 11:51 AM

Actually, both 48% and 84% are correct answers. Why?

Because there's something important .Roy didn't say: Do you proceed to try to get such "something" even if you do get it in the first try?

If you get the thing in the first try and you proceed, then the chance that you get the thing in exactly one try is 48% (0.6*0.4+0.4*0.6).

If you get the thing in the first try and you DON'T proceed to try again, then the chance is 84% (0.6+0.4*0.6)

It's my two cents.

I'm too lazy to draw the tree diagram...anyone?

Edit: My assumption is that: The event that you get the thing is independent i.e. the chance you get the thing is always 60%. I'm not talking about the case that there are 10 things, 6 of which are what you want (where you have 60% chance to get some thing you want)... just like what John says in the next post.

RE: Probobility Question? MATH by John Anderton on 02-14-2006 at 12:20 PM

Tbh the question isnt complete .... does the probability remain constant for both cases ??
If not then there are 2 options:

Is the option putback after you find that its wrong or is it removed ??
Like: there are 10 cards, 6 are red .... you need to pick out a red one .... you take one card and its black ... do you keep it aside and pick from the rest or put it back, reshuffle and try again ???

The other mistake is .... wait .... Mnjul already said it

quote:
Originally posted by Mnjul
Because there's something important .Roy didn't say: Do you proceed to try to get such "something" even if you do get it in the first try?

Assuming that its like my cards eg and you dont put it back and once you get the correct answer you quit (Mnjuls point)
Probability of getting it in the 1st attempt: 6/10 (Region A -> P(A))
Probability of getting it in the 2nd attempt: 6/9 (Region B -> P(B))

So to get the correct answer you want a region that contains a point in A and be but not in both of them (not in the intersection

)
P(AvB) => P (A union B)
P(A<>B) => P (A intersection B)
(I didnt want any confusion as to the symbolisms .... cause the actual ones arent available 8-)

P(A<>B) = (6/10) * (5/9) --> (Condition where it happens both the times)

= 1/3

Answer:
P(AvB) - P(A<>B)
= [P(A) + P(B) - P(A<>B)] - P(A<>B)

= P(A) + P(B) - 2 * P(A<>B)

= (6/10) + (5/9) - (2/3)

= 22/45

= 0.4889

= 48.89 %

Note that this is the correct answer (afaik) for the conditions: (Its been a year or so since i left probability

)

quote:
Originally posted by John Anderton in this post
Assuming that its like my cards eg and you dont put it back and once you get the correct answer you quit (Mnjuls point)

quote:
Originally posted by Mnjul
I'm too lazy to draw the tree diagram...anyone?

I was thinking about more of a regional diagram to support my answer but ill attatch them asap

It would be better if .Roy could tell us what he exactly wanted unless he wanted the answers in all aspects 8-)

RE: Probobility Question? MATH by qgroessl on 02-14-2006 at 01:17 PM

quote:
Originally posted by rix
Logically it would be 30%?! But heey, im not a rocket scientist.

That would be my guess... I'm too lazy to figure it out... and it makes a point... lol.

RE: Probobility Question? MATH by John Anderton on 02-14-2006 at 02:01 PM

quote:
Originally posted by groessl35

quote:
Originally posted by rix
Logically it would be 30%?! But heey, im not a rocket scientist.

That would be my guess... I'm too lazy to figure it out... and it makes a point... lol.

no way is it 30% if you knew how to find probability. Then again i think you are too young to learn that in class ?? I dont know ... i learnt it in year 12 ... our education system kinda sucks tbh

RE: Probobility Question? MATH by markee on 02-14-2006 at 04:09 PM

code:
You could also use binomial probability distribution functions or binomial cumulative distribution functions.

n = # or trials
p = P(success)
r = # of successes

in form of:
binompdf(n,p,r) or
binomcdf(n,p,r)

n = 2
p = 0.6 {decimal of 60%)

only 1 success

binompdf as only one point is wanted

r = 1

binompdf(2,0.6,1) = 0.48 = 48%

more than 1 success

binomcdf as multiple points are wanted (in this case binompdf would also work)

I will take the cumulative probability or all integers from before 1 and take this away from total probability (1 or 100%)

r=0 {the first integer preceeding 1}

1 - binomcdf(2,0.6,0) = 1 - 0.16 = 0.84 = 84%

Just another way of proving the answer but this is more simple (only takes a couple of numbers to enter and really basic maths) and will be more valuable later when dealing with more complex questions. Hope I helped and not confused you more, if you need more explaination just PM me.

quote:
Originally posted by rix
Logically it would be 30%?! But heey, im not a rocket scientist.

I can see that

*Markee runs

RE: Probobility Question? MATH by .Roy on 02-14-2006 at 07:57 PM

Thanks guys, forgive me for not wording my question exactly as I ment it.

I ment it happening AT LEAST once, the one with the 84%.

Thanks again

P.S I needed this for a certain game, and I wanted the thing to happen at least once, the more the marrier but i ended up getting them both to be the right one . Yay for 36%.

Oh and the probobility stays the same in all the tries. Meaning you dont take out anything from the "bag".