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Math Geeks UNITE!

I

Ilmater

Diamond Member
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 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 🙂
 
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.
Click to expand...
I'm 30, this isn't my homework. I'm not in school. I'm just trying to figure something out.
 
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.
 
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...
 
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...
Click to expand...

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.
 
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...
Click to expand...

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

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
 
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.
Click to expand...

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
 
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...
Click to expand...

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.
Click to expand...

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.
 
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 😀
 
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...
Click to expand...

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.
Click to expand...

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.
Click to expand...

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

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.
Click to expand...

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.
Click to expand...

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

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

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.
Click to expand...

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.
Click to expand...

😕 fail
 
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...
Click to expand...

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.
Click to expand...

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.
Click to expand...

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.
 
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.
Click to expand...

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 .....
 
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 .....
Click to expand...

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

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.
Click to expand...

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.
Click to expand...

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.
Click to expand...

You're right. I blew it.
 
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