Math Geeks UNITE!

Ilmater

Diamond Member
Jun 13, 2002
7,516
1
0
I'm trying to determine the probability that something happens after multiple chances. Each time, there is a 6.88% chance of success, and I can try up to 10 times. What is my chance of having a success after 10 tries. Is it...

X=P * Q^(n-1)

Where:

P = Probability of success
Q = Probability of failure
n = number of tries

Please tell me if I got that right.
 

The-Noid

Diamond Member
Nov 16, 2005
3,117
3
76
no it is 10 C 1 * (.0688^1) * (.9322^9) = .365738202

You did it a different way but that is the correct formula, as yours will only work if things are taken one time. When you need to know the chance of having 2 or more successes you need to tweak the formula.

C is for combination :)
 

Ilmater

Diamond Member
Jun 13, 2002
7,516
1
0
Originally posted by: Ilmater
I'm trying to determine the probability that something happens after multiple chances. Each time, there is a 6.88% chance of success, and I can try up to 10 times. What is my chance of having a success after 10 tries. Is it...

X=P * Q^(n-1)

Where:

P = Probability of success
Q = Probability of failure
n = number of tries

Please tell me if I got that right.
I'm 30, this isn't my homework. I'm not in school. I'm just trying to figure something out.
 

simpletron

Member
Oct 31, 2008
189
14
81
the easy way to do this problem, is figure out the probability that you lose on tries and subtract from 1, which is 1-(1-p)^n where p is the probability of success and n is the number of tries.

the other way is to take p*(1-p)^i and sum it over i = 0 to n-1. p*(1-p)^i is the odds of lossing the first i times and winning on the i+1 try. since you want to win within the first 10 try, you need to add up winning on the first, second, third,... tenth tries.

question: is it having one success or at least one success? because mine is for one or more whereas Yoxxy's is exactly one.
 

nickbits

Diamond Member
Mar 10, 2008
4,122
1
81
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...
 

AyashiKaibutsu

Diamond Member
Jan 24, 2004
9,306
3
81
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.
 

Q

Lifer
Jul 21, 2005
12,042
4
81
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

This is right, unless you are saying you have to get 3 heads in a row, then it's different.
 

Syringer

Lifer
Aug 2, 2001
19,333
2
71
Originally posted by: Ilmater
I'm trying to determine the probability that something happens after multiple chances. Each time, there is a 6.88% chance of success, and I can try up to 10 times. What is my chance of having a success after 10 tries. Is it...

X=P * Q^(n-1)

Where:

P = Probability of success
Q = Probability of failure
n = number of tries

Please tell me if I got that right.

If you want just one success, then basically you want the opposite of failing all the time.

So the simplest way to do it would be 1-Q^n = 1-(1-Q)^n = 1-(1-.0688)^10 = 0.509736
 

MrPickins

Diamond Member
May 24, 2003
9,117
766
126
Originally posted by: simpletron
the easy way to do this problem, is figure out the probability that you lose on tries and subtract from 1, which is 1-(1-p)^n where p is the probability of success and n is the number of tries.

the other way is to take p*(1-p)^i and sum it over i = 0 to n-1. p*(1-p)^i is the odds of lossing the first i times and winning on the i+1 try. since you want to win within the first 10 try, you need to add up winning on the first, second, third,... tenth tries.

question: is it having one success or at least one success? because mine is for one or more whereas Yoxxy's is exactly one.

This.

given n independent Bernoulli trials:

probability of all failures + probability of at least one success = 1

so,
1 - prob. of all fail = prob. of at least one success

1 - q^n = probability of at least one success in n trails


Edit: Syringer has the right answer
 

nickbits

Diamond Member
Mar 10, 2008
4,122
1
81
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.
 

Q

Lifer
Jul 21, 2005
12,042
4
81
I feel smart just for visiting this thread BTW -- I'm really not good at math but that is Statistics I believe and it's much easier :D
 

MrPickins

Diamond Member
May 24, 2003
9,117
766
126
Originally posted by: nickbits
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.

Take a class in probability/stats. You obviously haven't learned discrete probability distributions.
 

RedArmy

Platinum Member
Mar 1, 2005
2,648
0
0
Originally posted by: MrPickins
Originally posted by: nickbits
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.

Take a class in probability/stats. You obviously haven't learned discrete probability distributions.

I had a much longer reply for him but yours was more to the point.
 

Syringer

Lifer
Aug 2, 2001
19,333
2
71
Originally posted by: nickbits
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.

:confused: fail
 

Syringer

Lifer
Aug 2, 2001
19,333
2
71
Originally posted by: nickbits
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.

With four flips the possibilities are:

HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
THHT
TTTH
TTTT

Say the goal was to get at LEAST one instance of heads. There's only one instance where that doesn't occur, the last one..TTTT---therefore the odds of getting at least one heads is 15/16.
 

polarbear6

Golden Member
Jul 14, 2008
1,161
1
0
Originally posted by: Ilmater
I'm trying to determine the probability that something happens after multiple chances. Each time, there is a 6.88% chance of success, and I can try up to 10 times. What is my chance of having a success after 10 tries. Is it...

X=P * Q^(n-1)

Where:

P = Probability of success
Q = Probability of failure
n = number of tries

Please tell me if I got that right.

No... I dont remember probabilty probably ...
but i figure it has to be ..
p*q + p*q^2 + p*q^3 ....
some equation like that ...
I vaguely remember something like that .....
 

Legendary

Diamond Member
Jan 22, 2002
7,019
1
0
Chances of 1 success out of 10 = 100% - chances of 10 consecutive failures =
1 - ((1 - 0.0688)^10)
 

MrPickins

Diamond Member
May 24, 2003
9,117
766
126
Originally posted by: ganesh1
No... I dont remember probabilty probably ...
but i figure it has to be ..
p*q + p*q^2 + p*q^3 ....
some equation like that ...
I vaguely remember something like that .....

That way works, but finding the compliment is much easier.
 

JTsyo

Lifer
Nov 18, 2007
11,915
1,046
126
Originally posted by: Legendary
Chances of 1 success out of 10 = 100% - chances of 10 consecutive failures =
1 - ((1 - 0.0688)^10)

That's the way I usually do it.
 

nickbits

Diamond Member
Mar 10, 2008
4,122
1
81
Originally posted by: Syringer
Originally posted by: nickbits
Originally posted by: AyashiKaibutsu
Originally posted by: nickbits
The answer is 6.88%. The number of tries doesn't matter. It is like flipping a coin, the odds are always 50/50.

Unless you are asking something else...

You got it right on your second try : p

While it's true that for each time you flip the coin it's 50/50 no matter how many times you flip and what occured in those previous flips. What he's basicly asking is if you flip a coin twice what's the chance in that set of two flips atleast one heads will appear, which is not 50/50.

Yes it is. In 2 flips, 2 of 4 possilibites are heads. In 3 flips, 3 of 6 are heads. It is always 50/50.

With four flips the possibilities are:

HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
THHT
TTTH
TTTT

Say the goal was to get at LEAST one instance of heads. There's only one instance where that doesn't occur, the last one..TTTT---therefore the odds of getting at least one heads is 15/16.

You're right. I blew it.